Let L be the line through the intersection of the planes `3x-y+2z+1=0` and `3x-2y+z=3`. Then, the equation of the plane passing through `(2, 1, 4)` and perpendiculr to the line L is

Options

A. `x+y-z=2`
B. `x+y-z+1=0`
C. `x+y+z-7=0`
D. `2x-3y+4z=17`

Answer

To find the equation of the plane passing through the point (2, 1, 4) and perpendicular to the line L, which is the intersection of the two given planes, we can follow these steps:

### Step 1: Find the normal vectors of the given planes
The equations of the planes are:
1. \( P_1: 3x - y + 2z + 1 = 0 \)
2. \( P_2: 3x - 2y + z - 3 = 0 \)

The normal vector of plane \( P_1 \) is given by the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (3, -1, 2) \]

The normal vector of plane \( P_2 \) is:
\[ \mathbf{n_2} = (3, -2, 1) \]

### Step 2: Find the direction vector of line L
The direction vector of line L can be found by taking the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \):
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 2 \\ 3 & -2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{d} = \mathbf{i}((-1)(1) - (2)(-2)) - \mathbf{j}((3)(1) - (2)(3)) + \mathbf{k}((3)(-2) - (-1)(3)) \]
\[ = \mathbf{i}(-1 + 4) - \mathbf{j}(3 - 6) + \mathbf{k}(-6 + 3) \]
\[ = 3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k} \]
Thus, the direction vector of line L is:
\[ \mathbf{d} = (3, 3, -3) \]

### Step 3: Find the equation of the plane
The equation of a plane can be expressed as:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (a, b, c) \) is the normal vector to the plane.

Since the plane we are looking for is perpendicular to line L, we can use the direction vector \( \mathbf{d} = (3, 3, -3) \) as the normal vector of the plane. The point through which the plane passes is \( (2, 1, 4) \).

Thus, the equation becomes:
\[ 3(x - 2) + 3(y - 1) - 3(z - 4) = 0 \]

Expanding this:
\[ 3x - 6 + 3y - 3 - 3z + 12 = 0 \]
\[ 3x + 3y - 3z + 3 = 0 \]

Dividing through by 3 gives:
\[ x + y - z + 1 = 0 \]
or equivalently:
\[ x + y - z = -1 \]

### Final Answer
The equation of the plane passing through the point \( (2, 1, 4) \) and perpendicular to the line L is:
\[ x + y - z = -1 \]