Let the plane `P` pass through the intersection of the planes `2x + 3y – z = 2 and x + 2y + 3z = 6`, and be perpendicular to the plane `2x + y – z + 1 = 0`. If d is the distance of `P` from the point `(–7,1,1)`, then `d^(2)` is equal to

Options

A. `(25)/(83)`
B. `(250)/(83)`
C. `(15)/(53)`
D. `(250)/(82)`

Answer

To solve the problem, we need to find the equation of the plane \( P \) that passes through the intersection of two given planes and is perpendicular to a third plane. Then, we will calculate the distance from a specific point to this plane and find \( d^2 \).

### Step 1: Find the normal vectors of the given planes
The equations of the planes are:
1. \( 2x + 3y - z = 2 \) (Plane 1)
2. \( x + 2y + 3z = 6 \) (Plane 2)
3. \( 2x + y - z + 1 = 0 \) (Plane 3)

The normal vector of Plane 1, \( \mathbf{n_1} \), is \( (2, 3, -1) \).
The normal vector of Plane 2, \( \mathbf{n_2} \), is \( (1, 2, 3) \).
The normal vector of Plane 3, \( \mathbf{n_3} \), is \( (2, 1, -1) \).

### Step 2: Find the direction vector of the line of intersection of Plane 1 and Plane 2
To find the direction vector of the line of intersection of the two planes, we take the cross product of their normal vectors:

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

Calculating the determinant:

\[ \mathbf{d} = \mathbf{i}(3 \cdot 3 - (-1) \cdot 2) - \mathbf{j}(2 \cdot 3 - (-1) \cdot 1) + \mathbf{k}(2 \cdot 2 - 3 \cdot 1) \]
\[ = \mathbf{i}(9 + 2) - \mathbf{j}(6 + 1) + \mathbf{k}(4 - 3) \]
\[ = 11\mathbf{i} - 7\mathbf{j} + 1\mathbf{k} \]

Thus, the direction vector \( \mathbf{d} = (11, -7, 1) \).

### Step 3: Find the equation of the plane \( P \)
The plane \( P \) can be expressed in the form:

\[ P: 2x + 3y - z = 2 + \lambda (x + 2y + 3z - 6) \]

Expanding this gives:

\[ P: (2 + \lambda)x + (3 + 2\lambda)y + (-1 + 3\lambda)z - (2 + 6\lambda) = 0 \]

### Step 4: Ensure \( P \) is perpendicular to Plane 3
For \( P \) to be perpendicular to Plane 3, the dot product of their normal vectors must be zero:

\[ (2 + \lambda) \cdot 2 + (3 + 2\lambda) \cdot 1 + (-1 + 3\lambda)(-1) = 0 \]

Expanding this gives:

\[ 4 + 2\lambda + 3 + 2\lambda + 1 - 3\lambda = 0 \]
\[ 8 + \lambda = 0 \implies \lambda = -8 \]

### Step 5: Substitute \( \lambda \) back into the equation of plane \( P \)
Substituting \( \lambda = -8 \):

\[ P: (2 - 8)x + (3 - 16)y + (-1 - 24)z + (2 + 48) = 0 \]
\[ P: -6x - 13y - 25z + 46 = 0 \]
\[ 6x + 13y + 25z - 46 = 0 \]

### Step 6: Calculate the distance from the point \((-7, 1, 1)\) to the plane
The distance \( d \) from a point \((x_0, y_0, z_0)\) to the plane \( Ax + By + Cz + D = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, \( A = 6, B = 13, C = 25, D = -46 \), and the point is \((-7, 1, 1)\):

\[ d = \frac{|6(-7) + 13(1) + 25(1) - 46|}{\sqrt{6^2 + 13^2 + 25^2}} \]
\[ = \frac{|-42 + 13 + 25 - 46|}{\sqrt{36 + 169 + 625}} \]
\[ = \frac{|-42 + 38 - 46|}{\sqrt{830}} = \frac{|-50|}{\sqrt{830}} = \frac{50}{\sqrt{830}} \]

### Step 7: Calculate \( d^2 \)
Thus, \( d^2 = \left(\frac{50}{\sqrt{830}}\right)^2 = \frac{2500}{830} \).

### Final Answer
\[ d^2 = \frac{2500}{830} \]