Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S: `x+y+z=5.` If a line L passing through (1, -1, -1), parallel to the line PQ meets the plane Sat R, then `QR^2` is equal to :

Options

A. 2
B. 5
C. 7
D. 11

Answer

To solve the problem, we need to find the coordinates of the mirror image point \( Q \) of point \( P(1, 0, 1) \) with respect to the plane \( S: x + y + z = 5 \), determine the coordinates of point \( R \) where line \( L \) intersects the plane, and finally compute \( QR^2 \).

### Step 1: Find the coordinates of point \( Q \)

The formula for the mirror image of a point \( (x_1, y_1, z_1) \) with respect to the plane \( Ax + By + Cz + D = 0 \) is given by:

\[ \left( x', y', z' \right) = \left( x_1 - \frac{2A(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}, y_1 - \frac{2B(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}, z_1 - \frac{2C(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} \right) \]

For our plane \( S: x + y + z = 5 \), we can rewrite it as \( x + y + z - 5 = 0 \). Here, \( A = 1, B = 1, C = 1, D = -5 \).

Now, substituting \( P(1, 0, 1) \) into the formula:

1. Calculate \( Ax_1 + By_1 + Cz_1 + D \):
\[ 1 \cdot 1 + 1 \cdot 0 + 1 \cdot 1 - 5 = 1 + 0 + 1 - 5 = -3 \]

2. Calculate \( A^2 + B^2 + C^2 \):
\[ 1^2 + 1^2 + 1^2 = 3 \]

3. Now substitute into the formula:
\[ Q = \left( 1 - \frac{2 \cdot 1 \cdot (-3)}{3}, 0 - \frac{2 \cdot 1 \cdot (-3)}{3}, 1 - \frac{2 \cdot 1 \cdot (-3)}{3} \right) \]
\[ Q = \left( 1 + 2, 0 + 2, 1 + 2 \right) = (3, 2, 3) \]

### Step 2: Find the equation of line \( L \)

Line \( L \) passes through point \( (1, -1, -1) \) and is parallel to the line \( PQ \). The direction vector of line \( PQ \) can be found as follows:

\[ \text{Direction vector of } PQ = Q - P = (3 - 1, 2 - 0, 3 - 1) = (2, 2, 2) \]

The parametric equations of line \( L \) can be expressed as:
\[ x = 1 + 2t, \quad y = -1 + 2t, \quad z = -1 + 2t \]

### Step 3: Find the intersection point \( R \) with the plane \( S \)

To find point \( R \), we substitute the parametric equations into the plane equation \( x + y + z = 5 \):
\[ (1 + 2t) + (-1 + 2t) + (-1 + 2t) = 5 \]
\[ 2t + 2t + 2t - 1 = 5 \]
\[ 6t - 1 = 5 \implies 6t = 6 \implies t = 1 \]

Now, substituting \( t = 1 \) back into the parametric equations:
\[ x = 1 + 2(1) = 3, \quad y = -1 + 2(1) = 1, \quad z = -1 + 2(1) = 1 \]
Thus, \( R = (3, 1, 1) \).

### Step 4: Calculate \( QR^2 \)

Using the distance formula:
\[ QR^2 = (x_Q - x_R)^2 + (y_Q - y_R)^2 + (z_Q - z_R)^2 \]
Substituting the coordinates of \( Q(3, 2, 3) \) and \( R(3, 1, 1) \):
\[ QR^2 = (3 - 3)^2 + (2 - 1)^2 + (3 - 1)^2 = 0 + 1 + 4 = 5 \]

### Final Answer
Thus, the value of \( QR^2 \) is \( \boxed{5} \).