Let `P_(1):x-2y+3z=5 and P_(2):2x-3y+z+4=0` be two planes. The equation of the plane perpendicular to the line of intersection to the line of intersection of `P_(1)=0` and `P_(2)=` and passing through `(1,1,1)` is

Options

A. `11x-5y+7z-13=0`
B. `7x+5y+z=13`
C. `x+2y+z-4=0`
D. `x-2t+4z+3=0`

Answer

To find the equation of the plane that is perpendicular to the line of intersection of the two given planes \( P_1: x - 2y + 3z = 5 \) and \( P_2: 2x - 3y + z + 4 = 0 \) and passes through the point \( (1, 1, 1) \), we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \( Ax + By + Cz = D \) is represented by the coefficients \( (A, B, C) \).

For \( P_1: x - 2y + 3z = 5 \):
- The normal vector \( \mathbf{n_1} = (1, -2, 3) \).

For \( P_2: 2x - 3y + z + 4 = 0 \):
- The normal vector \( \mathbf{n_2} = (2, -3, 1) \).

### Step 2: Find a vector parallel to the line of intersection
The direction vector \( \mathbf{d} \) of the line of intersection of the two planes can be found using the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 2 & -3 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{d} = \mathbf{i}((-2)(1) - (3)(-3)) - \mathbf{j}((1)(1) - (3)(2)) + \mathbf{k}((1)(-3) - (-2)(2)) \]

\[ = \mathbf{i}(-2 + 9) - \mathbf{j}(1 - 6) + \mathbf{k}(-3 + 4) \]

\[ = 7\mathbf{i} + 5\mathbf{j} + 1\mathbf{k} \]

Thus, the direction vector \( \mathbf{d} = (7, 5, 1) \).

### Step 3: Find the normal vector of the required plane
The required plane is perpendicular to the line of intersection, which means its normal vector \( \mathbf{n} \) is parallel to the direction vector \( \mathbf{d} \). Therefore, we can take:

\[ \mathbf{n} = (7, 5, 1) \]

### Step 4: Write the equation of the plane
The equation of a plane with normal vector \( (a, b, c) \) passing through a point \( (x_0, y_0, z_0) \) is given by:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

Substituting \( a = 7, b = 5, c = 1 \) and the point \( (1, 1, 1) \):

\[ 7(x - 1) + 5(y - 1) + 1(z - 1) = 0 \]

Expanding this:

\[ 7x - 7 + 5y - 5 + z - 1 = 0 \]

Combining like terms:

\[ 7x + 5y + z - 13 = 0 \]

### Final Equation
Thus, the equation of the required plane is:

\[ 7x + 5y + z = 13 \]