OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Exercise 24 (D)

Question 1. Find the equation of the planes through the intersection of the planes \( x+2y+3z+ 5 = 0 \), \( 2x - 4y + z - 3 = 0 \) and the point \( (0, 1, 0) \).
Answer: The equation of a plane that passes through the line of intersection of two given planes, \( P_1 = 0 \) and \( P_2 = 0 \), is generally written as \( P_1 + \lambda P_2 = 0 \).
Given equations of planes are:
\( P_1: x+2y+3z+ 5 = 0 \) ...(1)
\( P_2: 2x - 4y + z - 3 = 0 \) ...(2)
The equation of the plane passing through the intersection of planes (1) and (2) is:
\( (x + 2y + 3z + 5) + \lambda(2x - 4y + z - 3) = 0 \) ...(3)
This plane also passes through the point \( (0, 1, 0) \). We substitute these coordinates into equation (3):
\( (0 + 2(1) + 3(0) + 5) + \lambda(2(0) - 4(1) + 0 - 3) = 0 \)
\( (2 + 5) + \lambda(-4 - 3) = 0 \)
\( 7 - 7\lambda = 0 \)
\( \implies 7\lambda = 7 \)
\( \implies \lambda = 1 \)
Now, we put the value of \( \lambda = 1 \) back into equation (3):
\( (x + 2y + 3z + 5) + 1(2x - 4y + z - 3) = 0 \)
\( x + 2y + 3z + 5 + 2x - 4y + z - 3 = 0 \)
\( \implies 3x - 2y + 4z + 2 = 0 \)
This is the required equation of the plane.
In simple words: To find this plane, we combine the two given plane equations with a special number called lambda (\( \lambda \)). Then, we use the given point to find what \( \lambda \) is. Once we know \( \lambda \), we put it back into the combined equation to get our final plane equation.

🎯 Exam Tip: Remember to substitute the given point into the combined plane equation carefully to correctly find the value of \( \lambda \). A small calculation error here will lead to a wrong final equation.

Question 2. Find the equation of the plane containing the line of intersection of the plane \( x + y + z - 6 = 0 \) and \( 2x + 3y + 4z + 5 = 0 \) passing through the point \( (1, 1, 1) \).
Answer: The equation of a plane passing through the line of intersection of two given planes can be written by combining their equations with a scalar parameter, \( \lambda \).
The given planes are:
\( x+y+z-6 = 0 \)
\( 2x+3y+4z+5 = 0 \)
In vector form, these can be written as:
\( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 6 = 0 \) and \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) + 5 = 0 \)
The equation of the plane passing through the line of intersection of these two planes is:
\( [\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 6] + \lambda[\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) + 5] = 0 \)
\( \implies \vec{r} \cdot [(\hat{i} + \hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})] - 6 + 5\lambda = 0 \)
\( \implies \vec{r} \cdot [(1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}] - 6 + 5\lambda = 0 \)
This new plane passes through the point \( (1, 1, 1) \), which means its position vector is \( \vec{r} = \hat{i} + \hat{j} + \hat{k} \). We substitute this into the equation:
\( (\hat{i} + \hat{j} + \hat{k}) \cdot [(1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}] - 6 + 5\lambda = 0 \)
\( \implies 1(1 + 2\lambda) + 1(1 + 3\lambda) + 1(1 + 4\lambda) - 6 + 5\lambda = 0 \)
\( \implies 1 + 2\lambda + 1 + 3\lambda + 1 + 4\lambda - 6 + 5\lambda = 0 \)
\( \implies 3 + 9\lambda - 6 + 5\lambda = 0 \)
\( \implies 14\lambda - 3 = 0 \)
\( \implies 14\lambda = 3 \)
\( \implies \lambda = \frac{3}{14} \)
Now, we substitute the value of \( \lambda \) back into the plane equation:
\( \vec{r} \cdot [(1 + 2(\frac{3}{14}))\hat{i} + (1 + 3(\frac{3}{14}))\hat{j} + (1 + 4(\frac{3}{14}))\hat{k}] - 6 + 5(\frac{3}{14}) = 0 \)
\( \implies \vec{r} \cdot [(\frac{14+6}{14})\hat{i} + (\frac{14+9}{14})\hat{j} + (\frac{14+12}{14})\hat{k}] - \frac{84}{14} + \frac{15}{14} = 0 \)
\( \implies \vec{r} \cdot [\frac{20}{14}\hat{i} + \frac{23}{14}\hat{j} + \frac{26}{14}\hat{k}] - \frac{69}{14} = 0 \)
Multiplying the entire equation by 14, we get:
\( \vec{r} \cdot [20\hat{i} + 23\hat{j} + 26\hat{k}] - 69 = 0 \)
This is the required equation of the plane.
In simple words: First, combine the two plane equations into one using a special number, \( \lambda \). Then, use the given point to figure out the value of \( \lambda \). Finally, put \( \lambda \) back into the combined equation to get the plane's answer in vector form.

🎯 Exam Tip: When dealing with vector equations, remember that the dot product of the position vector \( \vec{r} \) with the normal vector yields the plane equation. Ensure your arithmetic with fractions for \( \lambda \) is precise.

Question 3. Find the direction ratios of the normal to the plane passing through the point \( (2, 1, 3) \) and the line of intersection of the planes \( x + 2y + z = 3 \) and \( 2x - y - z = 5 \).
Answer: The direction ratios of the normal to a plane are simply the coefficients of \( x, y, \) and \( z \) in its Cartesian equation. We first need to find the equation of the plane.
The given planes are:
\( x+2y+z-3 = 0 \) ...(1)
\( 2x-y-z-5 = 0 \) ...(2)
The equation of the plane passing through the line of intersection of planes (1) and (2) is given by:
\( (x + 2y + z - 3) + \lambda(2x - y - z - 5) = 0 \) ...(3)
This plane passes through the point \( (2, 1, 3) \). We substitute these coordinates into equation (3):
\( (2 + 2(1) + 3 - 3) + \lambda(2(2) - 1 - 3 - 5) = 0 \)
\( (2 + 2 + 3 - 3) + \lambda(4 - 1 - 3 - 5) = 0 \)
\( \implies 4 + \lambda(-5) = 0 \)
\( \implies 4 - 5\lambda = 0 \)
\( \implies 5\lambda = 4 \)
\( \implies \lambda = \frac{4}{5} \)
Now, we substitute the value of \( \lambda = \frac{4}{5} \) back into equation (3):
\( (x + 2y + z - 3) + \frac{4}{5}(2x - y - z - 5) = 0 \)
To remove the fraction, multiply the entire equation by 5:
\( 5(x + 2y + z - 3) + 4(2x - y - z - 5) = 0 \)
\( 5x + 10y + 5z - 15 + 8x - 4y - 4z - 20 = 0 \)
\( \implies 13x + 6y + z - 35 = 0 \)
This is the required equation of the plane. The direction ratios of the normal to this plane are the coefficients of \( x, y, \) and \( z \).
Thus, the direction ratios of the normal to plane are \( <13, 6, 1> \).
In simple words: We first find the plane's full equation by mixing the two given planes with \( \lambda \) and using the point to solve for \( \lambda \). Once we have the plane's equation, the numbers in front of \( x, y, \) and \( z \) tell us the direction ratios of the line that is perpendicular to the plane.

🎯 Exam Tip: The direction ratios of the normal to a plane \( Ax+By+Cz+D=0 \) are \( \). Ensure all terms are on one side of the equation before identifying these ratios.

Question 4. Find the equation of the plane passing through the intersection of the planes \( \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) = 7 \) and \( \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9 \) and passing through the point \( (2, 1, 3) \).
Answer: To find the equation of a plane through the intersection of two given planes in vector form, we use the formula \( P_1 + \lambda P_2 = 0 \).
The given planes are:
\( \vec{r} \cdot (2\hat{i} + \hat{j} + 3\hat{k}) - 7 = 0 \)
\( \vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9 = 0 \)
The equation of the plane passing through the line of intersection of these two planes is:
\( [\vec{r} \cdot (2\hat{i} + \hat{j} + 3\hat{k}) - 7] + \lambda[\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0 \)
\( \implies \vec{r} \cdot [(2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 5\hat{j} + 3\hat{k})] - 7 - 9\lambda = 0 \)
\( \implies \vec{r} \cdot [(2 + 2\lambda)\hat{i} + (1 + 5\lambda)\hat{j} + (3 + 3\lambda)\hat{k}] - 7 - 9\lambda = 0 \)
This plane passes through the point \( (2, 1, 3) \), so its position vector is \( \vec{r} = 2\hat{i} + 1\hat{j} + 3\hat{k} \). Substitute this into the equation:
\( (2\hat{i} + \hat{j} + 3\hat{k}) \cdot [(2 + 2\lambda)\hat{i} + (1 + 5\lambda)\hat{j} + (3 + 3\lambda)\hat{k}] - 7 - 9\lambda = 0 \)
\( \implies 2(2 + 2\lambda) + 1(1 + 5\lambda) + 3(3 + 3\lambda) - 7 - 9\lambda = 0 \)
\( \implies 4 + 4\lambda + 1 + 5\lambda + 9 + 9\lambda - 7 - 9\lambda = 0 \)
\( \implies 14 + 9\lambda - 7 = 0 \)
\( \implies 7 + 9\lambda = 0 \)
\( \implies 9\lambda = -7 \)
\( \implies \lambda = -\frac{7}{9} \)
Now, we substitute the value of \( \lambda \) back into the plane equation:
\( \vec{r} \cdot [(2 + 2(-\frac{7}{9}))\hat{i} + (1 + 5(-\frac{7}{9}))\hat{j} + (3 + 3(-\frac{7}{9}))\hat{k}] - 7 - 9(-\frac{7}{9}) = 0 \)
\( \implies \vec{r} \cdot [(2 - \frac{14}{9})\hat{i} + (1 - \frac{35}{9})\hat{j} + (3 - \frac{21}{9})\hat{k}] - 7 + 7 = 0 \)
\( \implies \vec{r} \cdot [(\frac{18-14}{9})\hat{i} + (\frac{9-35}{9})\hat{j} + (\frac{27-21}{9})\hat{k}] = 0 \)
\( \implies \vec{r} \cdot [\frac{4}{9}\hat{i} - \frac{26}{9}\hat{j} + \frac{6}{9}\hat{k}] = 0 \)
Multiplying by 9, we get the equation:
\( \vec{r} \cdot [4\hat{i} - 26\hat{j} + 6\hat{k}] = 0 \)
We can also divide the normal vector by 2 for a simpler form, so the required equation is:
\( \vec{r} \cdot [2\hat{i} - 13\hat{j} + 3\hat{k}] = 0 \)
In simple words: We combine the vector equations of the two given planes using a \( \lambda \) term. Then, we plug in the coordinates of the given point into this new equation to find the value of \( \lambda \). Finally, substitute \( \lambda \) back to get the desired plane equation in vector form, representing all points on the plane.

🎯 Exam Tip: The vector equation of a plane is often given as \( \vec{r} \cdot \vec{N} = D \), where \( \vec{N} \) is the normal vector to the plane. When a plane passes through the origin, \( D \) is 0, so the equation becomes \( \vec{r} \cdot \vec{N} = 0 \).

Question 5. Find the equation of the plane passing through the intersection of the planes \( 2x + 3y - z + 1 = 0 \), \( x + y - 2z + 3 = 0 \) and perpendicular to the plane \( 3x - y - 2z - 4 = 0 \).
Answer: We need to find the equation of a plane that satisfies two conditions: passing through the intersection of two planes and being perpendicular to a third plane.
The given planes are:
\( 2x+3y-z+1 = 0 \)
\( x+y-2z+3 = 0 \)
The equation of a plane passing through the intersection of these two planes is:
\( (2x+3y-z+1) + \lambda(x+y-2z+3) = 0 \)
Rearranging the terms to group \( x, y, \) and \( z \):
\( (2+\lambda)x + (3+\lambda)y + (-1-2\lambda)z + (1+3\lambda) = 0 \) ...(1)
The normal vector to this plane is \( \vec{N_1} = <2+\lambda, 3+\lambda, -1-2\lambda> \).
The plane (1) is perpendicular to the plane \( 3x-y-2z-4 = 0 \).
The normal vector to this third plane is \( \vec{N_2} = <3, -1, -2> \).
If two planes are perpendicular, their normal vectors must be orthogonal (their dot product is zero).
So, \( \vec{N_1} \cdot \vec{N_2} = 0 \):
\( (2+\lambda)(3) + (3+\lambda)(-1) + (-1-2\lambda)(-2) = 0 \)
\( \implies 6 + 3\lambda - 3 - \lambda + 2 + 4\lambda = 0 \)
\( \implies 5 + 6\lambda = 0 \)
\( \implies 6\lambda = -5 \)
\( \implies \lambda = -\frac{5}{6} \)
Now, we substitute the value of \( \lambda = -\frac{5}{6} \) back into equation (1):
\( (2 - \frac{5}{6})x + (3 - \frac{5}{6})y + (-1 - 2(-\frac{5}{6}))z + (1 + 3(-\frac{5}{6})) = 0 \)
\( \implies (\frac{12-5}{6})x + (\frac{18-5}{6})y + (-1 + \frac{10}{6})z + (1 - \frac{15}{6}) = 0 \)
\( \implies \frac{7}{6}x + \frac{13}{6}y + (\frac{-6+10}{6})z + (\frac{6-15}{6}) = 0 \)
\( \implies \frac{7}{6}x + \frac{13}{6}y + \frac{4}{6}z - \frac{9}{6} = 0 \)
Multiplying the entire equation by 6 to clear the denominators:
\( 7x + 13y + 4z - 9 = 0 \)
This is the required equation of the plane.
In simple words: We first write the general equation for a plane passing through the crossing line of the first two planes, which includes a variable \( \lambda \). Then, because this new plane is perpendicular to a third plane, their "normal" lines (which are at 90 degrees to each plane) must also be at 90 degrees to each other. Using this rule, we find \( \lambda \) and then get the final plane equation.

🎯 Exam Tip: When a plane is perpendicular to another plane, the dot product of their normal vectors is zero. This condition is crucial for finding \( \lambda \).

Question 6. Find the equation of the plane through the point \( 2\hat{i} + \hat{j} - \hat{k} \) and passing through the line of intersection of the planes \( \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 0 \) and \( \vec{r} \cdot (\hat{j} + 2\hat{k}) = 0 \).
Answer: We will find the equation of the plane by combining the two given plane equations and then using the given point to solve for the unknown scalar.
The given planes are in vector form:
\( \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 0 \)
\( \vec{r} \cdot (\hat{j} + 2\hat{k}) = 0 \)
The equation of the plane passing through the line of intersection of these two planes is:
\( [\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k})] + \lambda[\vec{r} \cdot (\hat{j} + 2\hat{k})] = 0 \)
\( \implies \vec{r} \cdot [(\hat{i} + 3\hat{j} - \hat{k}) + \lambda(\hat{j} + 2\hat{k})] = 0 \)
\( \implies \vec{r} \cdot [\hat{i} + (3+\lambda)\hat{j} + (-1+2\lambda)\hat{k}] = 0 \) ...(1)
This plane passes through the point with position vector \( \vec{r} = 2\hat{i} + \hat{j} - \hat{k} \). We substitute this into equation (1):
\( (2\hat{i} + \hat{j} - \hat{k}) \cdot [\hat{i} + (3+\lambda)\hat{j} + (-1+2\lambda)\hat{k}] = 0 \)
\( \implies 2(1) + 1(3+\lambda) + (-1)(-1+2\lambda) = 0 \)
\( \implies 2 + 3 + \lambda + 1 - 2\lambda = 0 \)
\( \implies 6 - \lambda = 0 \)
\( \implies \lambda = 6 \)
Now, we substitute the value of \( \lambda = 6 \) back into equation (1):
\( \vec{r} \cdot [\hat{i} + (3+6)\hat{j} + (-1+2(6))\hat{k}] = 0 \)
\( \implies \vec{r} \cdot [\hat{i} + 9\hat{j} + (-1+12)\hat{k}] = 0 \)
\( \implies \vec{r} \cdot [\hat{i} + 9\hat{j} + 11\hat{k}] = 0 \)
This is the required equation of the plane.
In simple words: First, we blend the two given plane equations using a special number called \( \lambda \). Since the new plane must pass through a given point, we substitute that point's coordinates into our combined equation. This lets us find \( \lambda \), and then we plug \( \lambda \) back in to get the final answer.

🎯 Exam Tip: The position vector of a point \( (x, y, z) \) is \( x\hat{i} + y\hat{j} + z\hat{k} \). When a plane passes through a point, the coordinates of that point must satisfy the plane's equation.

Question 7. Find the vector equation of the plane through the point \( (1, 4, -2) \) and perpendicular to the line of intersection of the planes \( x + y + z = 10 \) and \( 2x - y + 3z = 18 \).
Answer: We need to find the equation of a plane that passes through a given point and is perpendicular to a specific line. This specific line is the intersection of two other planes.
The given planes are:
\( x+y+z-10 = 0 \)
\( 2x-y+3z-18 = 0 \)
The normal vectors to these planes are \( \vec{n_1} = <1, 1, 1> \) and \( \vec{n_2} = <2, -1, 3> \).
The line of intersection of these two planes is perpendicular to both \( \vec{n_1} \) and \( \vec{n_2} \). Therefore, the direction vector of this line of intersection, \( \vec{d} \), is given by the cross product of \( \vec{n_1} \) and \( \vec{n_2} \).
\( \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{vmatrix} \)
\( \implies \vec{d} = \hat{i}(1 \cdot 3 - 1 \cdot (-1)) - \hat{j}(1 \cdot 3 - 1 \cdot 2) + \hat{k}(1 \cdot (-1) - 1 \cdot 2) \)
\( \implies \vec{d} = \hat{i}(3+1) - \hat{j}(3-2) + \hat{k}(-1-2) \)
\( \implies \vec{d} = 4\hat{i} - \hat{j} - 3\hat{k} \)
The required plane is perpendicular to this line of intersection. This means that the direction vector of the line, \( \vec{d} \), serves as the normal vector \( \vec{N} \) for the new plane.
So, \( \vec{N} = 4\hat{i} - \hat{j} - 3\hat{k} \).
The plane passes through the point \( (1, 4, -2) \). Let this point be \( \vec{r_0} = \hat{i} + 4\hat{j} - 2\hat{k} \).
The vector equation of a plane passing through a point \( \vec{r_0} \) with normal vector \( \vec{N} \) is \( (\vec{r} - \vec{r_0}) \cdot \vec{N} = 0 \).
\( \implies \vec{r} \cdot \vec{N} = \vec{r_0} \cdot \vec{N} \)
\( \vec{r} \cdot (4\hat{i} - \hat{j} - 3\hat{k}) = (\hat{i} + 4\hat{j} - 2\hat{k}) \cdot (4\hat{i} - \hat{j} - 3\hat{k}) \)
\( \implies \vec{r} \cdot (4\hat{i} - \hat{j} - 3\hat{k}) = 1(4) + 4(-1) + (-2)(-3) \)
\( \implies \vec{r} \cdot (4\hat{i} - \hat{j} - 3\hat{k}) = 4 - 4 + 6 \)
\( \implies \vec{r} \cdot (4\hat{i} - \hat{j} - 3\hat{k}) = 6 \)
In Cartesian form, if \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), then \( 4x - y - 3z = 6 \).
This is the required vector equation of the plane.
In simple words: First, we find the direction of the line where the two given planes cross. We do this by taking the cross product of their normal lines. This direction then becomes the "normal" direction for our new plane. Finally, using this normal direction and the given point, we write down the plane's equation.

🎯 Exam Tip: The line of intersection of two planes is perpendicular to both of their normal vectors. Thus, its direction vector is given by the cross product of their normal vectors. This direction vector then serves as the normal to any plane perpendicular to it.

Question 8. Find the equation of the plane through the intersection of the planes \( x + 3y + 6 = 0 \) and \( 3x - y - 4z = 0 \) whose perpendicular distance from the origin is unity.
Answer: We need to find the equation of a plane that passes through the intersection of two planes and has a specific distance from the origin.
The given planes are:
\( x+3y+6 = 0 \)
\( 3x-y-4z = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (x+3y+6) + \lambda(3x-y-4z) = 0 \)
Rearranging the terms to group \( x, y, \) and \( z \):
\( (1+3\lambda)x + (3-\lambda)y + (-4\lambda)z + 6 = 0 \) ...(1)
The perpendicular distance of this plane from the origin \( (0, 0, 0) \) is unity (1). The formula for the distance from a point \( (x_0, y_0, z_0) \) to a plane \( Ax+By+Cz+D=0 \) is \( \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} \).
Here, \( (x_0, y_0, z_0) = (0, 0, 0) \), so the distance is \( \frac{|D|}{\sqrt{A^2+B^2+C^2}} \).
Given distance = 1:
\( \frac{|(1+3\lambda)(0) + (3-\lambda)(0) + (-4\lambda)(0) + 6|}{\sqrt{(1+3\lambda)^2 + (3-\lambda)^2 + (-4\lambda)^2}} = 1 \)
\( \implies \frac{|6|}{\sqrt{(1+3\lambda)^2 + (3-\lambda)^2 + 16\lambda^2}} = 1 \)
\( \implies 6 = \sqrt{(1+3\lambda)^2 + (3-\lambda)^2 + 16\lambda^2} \)
Squaring both sides:
\( 36 = (1+3\lambda)^2 + (3-\lambda)^2 + 16\lambda^2 \)
\( 36 = (1 + 6\lambda + 9\lambda^2) + (9 - 6\lambda + \lambda^2) + 16\lambda^2 \)
\( 36 = 1 + 6\lambda + 9\lambda^2 + 9 - 6\lambda + \lambda^2 + 16\lambda^2 \)
\( \implies 36 = 10 + 26\lambda^2 \)
\( \implies 26\lambda^2 = 26 \)
\( \implies \lambda^2 = 1 \)
\( \implies \lambda = \pm 1 \)
We have two possible values for \( \lambda \), so there will be two possible plane equations.
**Case 1: \( \lambda = 1 \)**
Substitute \( \lambda = 1 \) into equation (1):
\( (1+3(1))x + (3-1)y + (-4(1))z + 6 = 0 \)
\( \implies 4x + 2y - 4z + 6 = 0 \)
Dividing by 2:
\( \implies 2x + y - 2z + 3 = 0 \)
**Case 2: \( \lambda = -1 \)**
Substitute \( \lambda = -1 \) into equation (1):
\( (1+3(-1))x + (3-(-1))y + (-4(-1))z + 6 = 0 \)
\( \implies (1-3)x + (3+1)y + (4)z + 6 = 0 \)
\( \implies -2x + 4y + 4z + 6 = 0 \)
Dividing by -2:
\( \implies x - 2y - 2z - 3 = 0 \)
These are the two required equations of the planes.
In simple words: First, we form a general plane equation from the intersection of the two given planes, which involves a variable \( \lambda \). We then use the rule for finding the distance from a point (the origin, in this case) to a plane. We set this distance to 1 and solve for \( \lambda \). Since we get two possible values for \( \lambda \), there are two different planes that fit the conditions.

🎯 Exam Tip: Remember that taking the square root can lead to both positive and negative values, so always consider both \( \pm \lambda \) to find all possible solutions. Also, ensure the distance formula is applied correctly, especially for the origin.

Question 9. Find the equation of the plane passing through the intersection of the planes \( 4x - y + z = 10 \) and \( x + y - z = 4 \) and parallel to the line with direction ratios \( 2,1,1 \). Also, find the perpendicular distance of \( (1, 1, 1) \) from this plane.
Answer: We need to find the equation of a plane that satisfies two conditions: passing through the intersection of two planes and being parallel to a given line. Then we'll find a distance.
The given planes are:
\( 4x-y+z-10 = 0 \)
\( x+y-z-4 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (4x-y+z-10) + \lambda(x+y-z-4) = 0 \)
Rearranging terms to group \( x, y, \) and \( z \):
\( (4+\lambda)x + (-1+\lambda)y + (1-\lambda)z + (-10-4\lambda) = 0 \) ...(1)
The normal vector to this plane is \( \vec{N} = <4+\lambda, -1+\lambda, 1-\lambda> \).
The plane (1) is parallel to the line with direction ratios \( <2, 1, 1> \). Let this line's direction vector be \( \vec{d} = <2, 1, 1> \).
If a plane is parallel to a line, its normal vector is perpendicular to the line's direction vector (their dot product is zero).
So, \( \vec{N} \cdot \vec{d} = 0 \):
\( (4+\lambda)(2) + (-1+\lambda)(1) + (1-\lambda)(1) = 0 \)
\( \implies 8 + 2\lambda - 1 + \lambda + 1 - \lambda = 0 \)
\( \implies 8 + 2\lambda = 0 \)
\( \implies 2\lambda = -8 \)
\( \implies \lambda = -4 \)
Now, we substitute the value of \( \lambda = -4 \) back into equation (1):
\( (4+(-4))x + (-1+(-4))y + (1-(-4))z + (-10-4(-4)) = 0 \)
\( \implies 0x + (-5)y + (5)z + (-10+16) = 0 \)
\( \implies -5y + 5z + 6 = 0 \)
Multiplying by -1, the equation of the required plane is:
\( 5y - 5z - 6 = 0 \)
Now, we need to find the perpendicular distance of the point \( (1, 1, 1) \) from this plane \( 5y - 5z - 6 = 0 \).
Using the distance formula \( \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} \):
Here, \( A=0, B=5, C=-5, D=-6 \), and the point is \( (x_0, y_0, z_0) = (1, 1, 1) \).
Distance \( = \frac{|0(1) + 5(1) - 5(1) - 6|}{\sqrt{0^2 + 5^2 + (-5)^2}} \)
\( = \frac{|0 + 5 - 5 - 6|}{\sqrt{0 + 25 + 25}} \)
\( = \frac{|-6|}{\sqrt{50}} \)
\( = \frac{6}{\sqrt{25 \cdot 2}} \)
\( = \frac{6}{5\sqrt{2}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{2} \):
\( = \frac{6\sqrt{2}}{5\sqrt{2} \cdot \sqrt{2}} = \frac{6\sqrt{2}}{10} = \frac{3\sqrt{2}}{5} \) units.
In simple words: First, we use a \( \lambda \) term to create a general equation for the plane that passes through the intersection of the two given planes. Because this new plane is parallel to a line, its normal line must be perpendicular to that line. This helps us find \( \lambda \). With the plane's equation, we then use a special formula to measure the straight-line distance from a specific point to that plane.

🎯 Exam Tip: Remember that if a plane is parallel to a line, their normal vector and direction vector respectively are orthogonal. Also, correctly rationalize denominators when finding distances.

Question 10. Find the equation of the plane which passes through the line of intersection of the planes \( x + 5y - 2z = 6 \) and \( 5x - 4y + 5z = 2 \) and parallel to the line joining the points \( (5, 1, 4) \) and \( (4, -2, 3) \).
Answer: We need to determine the equation of a plane given its relationship with two intersecting planes and a line formed by two points.
The given planes are:
\( x+5y-2z-6 = 0 \)
\( 5x-4y+5z-2 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (x+5y-2z-6) + \lambda(5x-4y+5z-2) = 0 \)
Rearranging terms:
\( (1+5\lambda)x + (5-4\lambda)y + (-2+5\lambda)z + (-6-2\lambda) = 0 \) ...(1)
The normal vector to this plane is \( \vec{N} = <1+5\lambda, 5-4\lambda, -2+5\lambda> \).
The plane (1) is parallel to the line joining points \( P_1(5, 1, 4) \) and \( P_2(4, -2, 3) \).
The direction vector of this line, \( \vec{d} \), is found by subtracting the coordinates of the points:
\( \vec{d} = <4-5, -2-1, 3-4> = <-1, -3, -1> \)
We can also use the direction ratios \( <1, 3, 1> \) (by multiplying by -1, which only changes the direction but not the line itself).
Since the plane is parallel to the line, its normal vector \( \vec{N} \) must be perpendicular to the line's direction vector \( \vec{d} \). Therefore, their dot product is zero.
\( \vec{N} \cdot \vec{d} = 0 \)
Using \( \vec{d} = <1, 3, 1> \):
\( (1+5\lambda)(1) + (5-4\lambda)(3) + (-2+5\lambda)(1) = 0 \)
\( \implies 1 + 5\lambda + 15 - 12\lambda - 2 + 5\lambda = 0 \)
\( \implies (5 - 12 + 5)\lambda + (1 + 15 - 2) = 0 \)
\( \implies -2\lambda + 14 = 0 \)
\( \implies 2\lambda = 14 \)
\( \implies \lambda = 7 \)
Now, we substitute the value of \( \lambda = 7 \) back into equation (1):
\( (1+5(7))x + (5-4(7))y + (-2+5(7))z + (-6-2(7)) = 0 \)
\( \implies (1+35)x + (5-28)y + (-2+35)z + (-6-14) = 0 \)
\( \implies 36x - 23y + 33z - 20 = 0 \)
This is the required equation of the plane.
In simple words: First, we form a general equation for the plane using the crossing line of the first two planes and a variable \( \lambda \). Then, we find the direction of the second line using the two given points. Because the new plane is parallel to this second line, its normal direction must be perpendicular to the line's direction. This allows us to find \( \lambda \), and then we plug \( \lambda \) back in to get the final plane equation.

🎯 Exam Tip: The direction ratios of a line passing through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are \( \). Remember this for vector operations.

Question 11. Find the equation to the plane through the line of intersection of the planes \( ax + by +cz + d = 0 \), \( a'x + b'y + c'z + d' = 0 \) and
(i) parallel to the x-axis,
(ii) perpendicular to xy-plane.
Answer: We need to find the equation of a plane that passes through the intersection of two given planes, under two different conditions.
The given planes are:
\( ax+by+cz+d = 0 \)
\( a'x+b'y+c'z+d' = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (ax+by+cz+d) + \lambda(a'x+b'y+c'z+d') = 0 \)
Rearranging terms to group \( x, y, \) and \( z \):
\( (a+\lambda a')x + (b+\lambda b')y + (c+\lambda c')z + (d+\lambda d') = 0 \) ...(1)
The normal vector to this plane is \( \vec{N} = \).

**(i) Parallel to the x-axis**
If the plane (1) is parallel to the x-axis, its normal vector \( \vec{N} \) must be perpendicular to the direction vector of the x-axis. The direction vector of the x-axis is \( \hat{i} = <1, 0, 0> \).
So, \( \vec{N} \cdot \hat{i} = 0 \):
\( (a+\lambda a')(1) + (b+\lambda b')(0) + (c+\lambda c')(0) = 0 \)
\( \implies a+\lambda a' = 0 \)
\( \implies \lambda a' = -a \)
\( \implies \lambda = -\frac{a}{a'} \)
Substitute this value of \( \lambda \) back into equation (1):
\( (a - \frac{a}{a'}a')x + (b - \frac{a}{a'}b')y + (c - \frac{a}{a'}c')z + (d - \frac{a}{a'}d') = 0 \)
\( \implies 0x + (\frac{a'b - ab'}{a'})y + (\frac{a'c - ac'}{a'})z + (\frac{a'd - ad'}{a'}) = 0 \)
Multiplying by \( a' \) to remove the denominator:
\( \implies (a'b - ab')y + (a'c - ac')z + (a'd - ad') = 0 \)
This can also be written as \( a'(by + cz + d) = a(b'y + c'z + d') \), which is the required equation of the plane.

**(ii) Perpendicular to the xy-plane**
The xy-plane has the equation \( z=0 \), and its normal vector is \( \vec{N}_{xy} = <0, 0, 1> \).
If plane (1) is perpendicular to the xy-plane, then their normal vectors must be orthogonal (their dot product is zero).
So, \( \vec{N} \cdot \vec{N}_{xy} = 0 \):
\( (a+\lambda a')(0) + (b+\lambda b')(0) + (c+\lambda c')(1) = 0 \)
\( \implies c+\lambda c' = 0 \)
\( \implies \lambda c' = -c \)
\( \implies \lambda = -\frac{c}{c'} \)
Substitute this value of \( \lambda \) back into equation (1):
\( (a - \frac{c}{c'}a')x + (b - \frac{c}{c'}b')y + (c - \frac{c}{c'}c')z + (d - \frac{c}{c'}d') = 0 \)
\( \implies (\frac{ac' - a'c}{c'})x + (\frac{bc' - b'c}{c'})y + 0z + (\frac{dc' - d'c}{c'}) = 0 \)
Multiplying by \( c' \) to remove the denominator:
\( \implies (ac' - a'c)x + (bc' - b'c)y + (dc' - d'c) = 0 \)
This can also be written as \( c'(ax + by + d) = c(a'x + b'y + d') \), which is the required equation of the plane.
In simple words: First, we write a general equation for a plane passing through the intersection of the two given planes, introducing a \( \lambda \) variable. For part (i), if the plane is parallel to the x-axis, its normal line must be perpendicular to the x-axis. This helps us find \( \lambda \). For part (ii), if the plane is perpendicular to the xy-plane, its normal line must be perpendicular to the normal line of the xy-plane. This gives us a different \( \lambda \). In both cases, we substitute \( \lambda \) back to get the final plane equation.

🎯 Exam Tip: Remember that a plane parallel to an axis has its normal vector perpendicular to that axis's direction vector. A plane perpendicular to another plane has their normal vectors orthogonal to each other.

Question 12. Find the equation of the plane which passes through the y-axis and the point \( (4, 2, -3) \).
Answer: A plane that passes through an axis has a special form. If a plane passes through the y-axis, it must contain all points \( (0, y, 0) \), which means it passes through the origin \( (0, 0, 0) \).
Such a plane contains the y-axis, so its equation will not have a \( y \) term, and no constant term if it also passes through the origin. Therefore, it can be written as \( Ax + Cz = 0 \) or, more simply, by combining the \( x=0 \) (yz-plane) and \( z=0 \) (xy-plane) equations with a parameter.
The equation of a plane passing through the y-axis can be expressed as:
\( z + kx = 0 \) (or \( x + kz = 0 \), both forms are valid) ...(1)
The plane also passes through the point \( (4, 2, -3) \). We substitute these coordinates into equation (1). Note that the \( y \)-coordinate does not affect the equation since the plane contains the entire y-axis.
\( -3 + k(4) = 0 \)
\( \implies -3 + 4k = 0 \)
\( \implies 4k = 3 \)
\( \implies k = \frac{3}{4} \)
Now, we substitute the value of \( k = \frac{3}{4} \) back into equation (1):
\( z + \frac{3}{4}x = 0 \)
Multiplying by 4 to remove the denominator:
\( \implies 4z + 3x = 0 \)
Rearranging the terms, the required equation of the plane is:
\( 3x + 4z = 0 \)
In simple words: When a plane contains the y-axis, its equation takes a simpler form, like \( z + kx = 0 \), because it passes through the origin and covers all points along the y-axis. We then use the given point to find the value of \( k \), and plug it back in to get the final plane equation.

🎯 Exam Tip: If a plane passes through an axis (e.g., y-axis), its general equation can be simplified. For example, for the y-axis, the equation will not depend on the y-coordinate directly, and can be written as \( Ax+Cz=0 \) or \( x+\lambda z=0 \).

Question 13. Find the equation of the plane passing through the line of intersection of the planes \( x + 2y + 3z - 5 = 0 \) and \( 3x - 2y - z + 1 = 0 \) and cutting off equal intercepts on x-axis and z-axis.
Answer: We need to find a plane that passes through the line where two planes meet and has equal segments cut off on the x-axis and z-axis.
The given planes are:
\( x+2y+3z-5 = 0 \)
\( 3x-2y-z+1 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (x+2y+3z-5) + \lambda(3x-2y-z+1) = 0 \)
Rearranging terms to group \( x, y, \) and \( z \):
\( (1+3\lambda)x + (2-2\lambda)y + (3-\lambda)z + (-5+\lambda) = 0 \)
Moving the constant term to the right side:
\( (1+3\lambda)x + (2-2\lambda)y + (3-\lambda)z = 5-\lambda \) ...(1)
To find the intercepts, we divide the entire equation by the constant term on the right side to get the intercept form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
The x-intercept \( X_i = \frac{5-\lambda}{1+3\lambda} \).
The z-intercept \( Z_i = \frac{5-\lambda}{3-\lambda} \).
We are given that the plane cuts off equal intercepts on the x-axis and z-axis, so \( X_i = Z_i \):
\( \frac{5-\lambda}{1+3\lambda} = \frac{5-\lambda}{3-\lambda} \)
This equation holds if either \( 5-\lambda = 0 \) or \( 1+3\lambda = 3-\lambda \).
**Case 1: \( 5-\lambda = 0 \)**
\( \implies \lambda = 5 \)
If \( \lambda = 5 \), the plane equation becomes:
\( (1+3(5))x + (2-2(5))y + (3-5)z = 5-5 \)
\( \implies 16x - 8y - 2z = 0 \)
Dividing by 2: \( 8x - 4y - z = 0 \). This plane passes through the origin, meaning its intercepts are all zero, which are "equal". While mathematically correct, often "equal intercepts" implies non-zero intercepts. Following the source, we will consider the second case.
**Case 2: \( 1+3\lambda = 3-\lambda \)** (assuming \( 5-\lambda \neq 0 \))
\( \implies 4\lambda = 2 \)
\( \implies \lambda = \frac{2}{4} = \frac{1}{2} \)
Now, we substitute the value of \( \lambda = \frac{1}{2} \) back into equation (1):
\( (1+3(\frac{1}{2}))x + (2-2(\frac{1}{2}))y + (3-\frac{1}{2})z = 5-\frac{1}{2} \)
\( \implies (\frac{2+3}{2})x + (2-1)y + (\frac{6-1}{2})z = \frac{10-1}{2} \)
\( \implies \frac{5}{2}x + 1y + \frac{5}{2}z = \frac{9}{2} \)
Multiplying the entire equation by 2 to clear denominators:
\( 5x + 2y + 5z = 9 \)
\( \implies 5x + 2y + 5z - 9 = 0 \)
This is the required equation of the plane.
In simple words: We first set up a general plane equation that includes a \( \lambda \) variable and passes through the crossing line of the two given planes. Then, we look at the parts of this equation that tell us where the plane cuts the x-axis and z-axis. By setting these two parts equal, we find the value of \( \lambda \). Finally, we plug \( \lambda \) back into the general equation to get the final plane.

🎯 Exam Tip: When setting intercepts equal, consider both possibilities: when the numerator is zero (plane passes through origin) and when the denominators are equal. Context often implies non-zero intercepts, guiding the choice of \( \lambda \).

Question 14. Find the vector equation of the plane containing the line of intersection of the planes \( x - 3y + 2z - 5 = 0 \) and \( 2x - y + 3z - 1 = 0 \) and passing through the point \( (1, -2, 3) \).
Answer: We will find the equation of the plane by first combining the equations of the two given planes with a \( \lambda \) parameter, and then using the given point to solve for \( \lambda \).
The given planes are:
\( x - 3y + 2z - 5 = 0 \)
\( 2x - y + 3z - 1 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (x - 3y + 2z - 5) + \lambda(2x - y + 3z - 1) = 0 \)
Rearranging terms to group \( x, y, \) and \( z \):
\( (1+2\lambda)x + (-3-\lambda)y + (2+3\lambda)z + (-5-\lambda) = 0 \) ...(1)
This plane passes through the point \( (1, -2, 3) \). We substitute these coordinates into equation (1):
\( (1+2\lambda)(1) + (-3-\lambda)(-2) + (2+3\lambda)(3) + (-5-\lambda) = 0 \)
\( \implies 1 + 2\lambda + 6 + 2\lambda + 6 + 9\lambda - 5 - \lambda = 0 \)
\( \implies (2+2+9-1)\lambda + (1+6+6-5) = 0 \)
\( \implies 12\lambda + 8 = 0 \)
\( \implies 12\lambda = -8 \)
\( \implies \lambda = -\frac{8}{12} = -\frac{2}{3} \)
Now, we substitute the value of \( \lambda = -\frac{2}{3} \) back into equation (1):
\( (1+2(-\frac{2}{3}))x + (-3-(-\frac{2}{3}))y + (2+3(-\frac{2}{3}))z + (-5-(-\frac{2}{3})) = 0 \)
\( \implies (1-\frac{4}{3})x + (-3+\frac{2}{3})y + (2-2)z + (-5+\frac{2}{3}) = 0 \)
\( \implies (\frac{3-4}{3})x + (\frac{-9+2}{3})y + 0z + (\frac{-15+2}{3}) = 0 \)
\( \implies -\frac{1}{3}x - \frac{7}{3}y + 0z - \frac{13}{3} = 0 \)
Multiplying the entire equation by -3 to clear denominators and make the leading term positive:
\( x + 7y + 13 = 0 \)
This is the Cartesian equation of the plane. To convert it to vector form, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 7\hat{j} + 0\hat{k}) + 13 = 0 \)
\( \implies \vec{r} \cdot (\hat{i} + 7\hat{j}) + 13 = 0 \)
This is the required vector equation of the plane.
In simple words: We first combine the two plane equations using a special number \( \lambda \). Then, we use the given point to find the exact value of \( \lambda \). Once we have \( \lambda \), we put it back into the combined equation to get the plane's equation. Finally, we change this equation into its vector form.

🎯 Exam Tip: The vector equation of a plane \( Ax+By+Cz+D=0 \) is \( \vec{r} \cdot (A\hat{i}+B\hat{j}+C\hat{k}) + D = 0 \). The coefficients of \( x, y, z \) form the normal vector.

Question 15. Find the equation of the plane passing through the line of intersection of the planes \( 2x + y - z = 3 \) and \( 5x - 3y + 4z = 9 \) and parallel to the line \( \frac{x-1}{2} = \frac{y-3}{4} = \frac{z-5}{5} \).
Answer: We are looking for the equation of a plane that passes through the intersection of two given planes and is parallel to a specified line.
The given planes are:
\( 2x+y-z-3 = 0 \)
\( 5x-3y+4z-9 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is:
\( (2x+y-z-3) + \lambda(5x-3y+4z-9) = 0 \)
Rearranging terms to group \( x, y, \) and \( z \):
\( (2+5\lambda)x + (1-3\lambda)y + (-1+4\lambda)z + (-3-9\lambda) = 0 \) ...(1)
The normal vector to this plane is \( \vec{N} = <2+5\lambda, 1-3\lambda, -1+4\lambda> \).
The plane (1) is parallel to the line \( \frac{x-1}{2} = \frac{y-3}{4} = \frac{z-5}{5} \).
The direction vector of this line is \( \vec{d} = <2, 4, 5> \).
If a plane is parallel to a line, its normal vector must be perpendicular to the line's direction vector (their dot product is zero).
So, \( \vec{N} \cdot \vec{d} = 0 \):
\( (2+5\lambda)(2) + (1-3\lambda)(4) + (-1+4\lambda)(5) = 0 \)
\( \implies 4 + 10\lambda + 4 - 12\lambda - 5 + 20\lambda = 0 \)
\( \implies (10 - 12 + 20)\lambda + (4 + 4 - 5) = 0 \)
\( \implies 18\lambda + 3 = 0 \)
\( \implies 18\lambda = -3 \)
\( \implies \lambda = -\frac{3}{18} = -\frac{1}{6} \)
Now, we substitute the value of \( \lambda = -\frac{1}{6} \) back into equation (1):
\( (2+5(-\frac{1}{6}))x + (1-3(-\frac{1}{6}))y + (-1+4(-\frac{1}{6}))z + (-3-9(-\frac{1}{6})) = 0 \)
\( \implies (2-\frac{5}{6})x + (1+\frac{3}{6})y + (-1-\frac{4}{6})z + (-3+\frac{9}{6}) = 0 \)
\( \implies (\frac{12-5}{6})x + (\frac{6+3}{6})y + (\frac{-6-4}{6})z + (\frac{-18+9}{6}) = 0 \)
\( \implies \frac{7}{6}x + \frac{9}{6}y - \frac{10}{6}z - \frac{9}{6} = 0 \)
Multiplying the entire equation by 6 to clear the denominators:
\( 7x + 9y - 10z - 9 = 0 \)
This is the required equation of the plane.
In simple words: First, we form a general equation for the plane using the crossing point of the first two planes and a \( \lambda \) variable. We also find the direction of the given line. Since the new plane is parallel to this line, its "normal" direction must be perpendicular to the line's direction. This rule helps us find \( \lambda \), which we then use to write the final plane equation.

🎯 Exam Tip: Always remember that when a plane is parallel to a line, its normal vector is perpendicular to the line's direction vector. This implies their dot product is zero, which is a key step in solving for \( \lambda \).

Question 16. Show that the line \( \frac{x+4}{3} = \frac{y+6}{5} = \frac{z-1}{-2} \) and the planes \( 3x - 2y + z + 5 = 0 \) and \( 2x + 3y + 4z - 4 = 0 \) intersect. Find the equation of the plane in which they lie and also their point of intersection.
Answer: This question asks for three things: proving intersection, finding the point of intersection, and finding the equation of the plane containing them.
The given line (1) is \( \frac{x+4}{3} = \frac{y+6}{5} = \frac{z-1}{-2} \). Let this equal \( t \).
Any point on the line (1) can be written as \( (x, y, z) = (3t-4, 5t-6, -2t+1) \).
The given planes are:
\( P_2: 3x-2y+z+5 = 0 \) ...(2)
\( P_3: 2x+3y+4z-4 = 0 \) ...(3)

**Part 1: Show that the line intersects the planes**
For the line to intersect the planes, a point on the line must satisfy both plane equations. Substitute the coordinates of the point on the line into each plane equation:
For Plane (2):
\( 3(3t-4) - 2(5t-6) + (-2t+1) + 5 = 0 \)
\( \implies 9t - 12 - 10t + 12 - 2t + 1 + 5 = 0 \)
\( \implies (9 - 10 - 2)t + (-12 + 12 + 1 + 5) = 0 \)
\( \implies -3t + 6 = 0 \)
\( \implies 3t = 6 \)
\( \implies t = 2 \)
For Plane (3):
\( 2(3t-4) + 3(5t-6) + 4(-2t+1) - 4 = 0 \)
\( \implies 6t - 8 + 15t - 18 - 8t + 4 - 4 = 0 \)
\( \implies (6 + 15 - 8)t + (-8 - 18 + 4 - 4) = 0 \)
\( \implies 13t - 26 = 0 \)
\( \implies 13t = 26 \)
\( \implies t = 2 \)
Since the same value of \( t=2 \) satisfies both plane equations, the line intersects both planes at a single point. Thus, the line and planes intersect.

**Part 2: Find the point of intersection**
Substitute \( t=2 \) back into the point coordinates for the line:
\( x = 3(2) - 4 = 6 - 4 = 2 \)
\( y = 5(2) - 6 = 10 - 6 = 4 \)
\( z = -2(2) + 1 = -4 + 1 = -3 \)
So, the point of intersection is \( (2, 4, -3) \).

**Part 3: Find the equation of the plane in which they lie**
The required plane contains the line of intersection of planes (2) and (3) and also contains the given line (1).
The equation of a plane passing through the line of intersection of planes (2) and (3) is:
\( (3x-2y+z+5) + \lambda(2x+3y+4z-4) = 0 \)
Rearranging terms:
\( (3+2\lambda)x + (-2+3\lambda)y + (1+4\lambda)z + (5-4\lambda) = 0 \) ...(4)
The normal vector to this plane is \( \vec{N_4} = <3+2\lambda, -2+3\lambda, 1+4\lambda> \).
Since the plane (4) contains line (1), its normal vector \( \vec{N_4} \) must be perpendicular to the direction vector of line (1). The direction vector of line (1) is \( \vec{d_1} = <3, 5, -2> \).
So, \( \vec{N_4} \cdot \vec{d_1} = 0 \):
\( (3+2\lambda)(3) + (-2+3\lambda)(5) + (1+4\lambda)(-2) = 0 \)
\( \implies 9 + 6\lambda - 10 + 15\lambda - 2 - 8\lambda = 0 \)
\( \implies (6+15-8)\lambda + (9-10-2) = 0 \)
\( \implies 13\lambda - 3 = 0 \)
\( \implies 13\lambda = 3 \)
\( \implies \lambda = \frac{3}{13} \)
Now, we substitute the value of \( \lambda = \frac{3}{13} \) back into equation (4):
\( (3+2(\frac{3}{13}))x + (-2+3(\frac{3}{13}))y + (1+4(\frac{3}{13}))z + (5-4(\frac{3}{13})) = 0 \)
\( \implies (\frac{39+6}{13})x + (\frac{-26+9}{13})y + (\frac{13+12}{13})z + (\frac{65-12}{13}) = 0 \)
\( \implies \frac{45}{13}x - \frac{17}{13}y + \frac{25}{13}z + \frac{53}{13} = 0 \)
Multiplying the entire equation by 13 to clear the denominators:
\( 45x - 17y + 25z + 53 = 0 \)
This is the required equation of the plane containing the given line and the intersection of the two planes.
In simple words: First, we check if the line crosses both planes by substituting a general point from the line into each plane equation. If we get the same \( t \) value, they intersect. This \( t \) value also helps us find the exact point where they cross. Next, to find the plane that holds everything, we use a general equation for a plane passing through the intersection of the two given planes. Since this new plane also contains the given line, its normal line must be perpendicular to the given line's direction. This allows us to find the final equation of the plane.

🎯 Exam Tip: To show a line intersects a plane, substitute the parametric form of the line into the plane's equation. If a consistent value for the parameter (e.g., \( t \)) is found, an intersection exists. If a plane contains a line, the normal to the plane is perpendicular to the direction vector of the line.