Let P be the image of the point `(3, 1, 7)` with respect to the plane `x-y+z=3`. Then, the equation of the plane passing through P and containing the straight line `(x)/(1)=(y)/(2)=(z)/(1)` is

Options

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

Answer

To solve the problem, we need to find the image of the point \( P(3, 1, 7) \) with respect to the plane given by the equation \( x - y + z = 3 \). Then, we will find the equation of the plane that passes through this image point \( P \) and contains the straight line defined by the symmetric equations \( \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \).

### Step 1: Find the image of point \( P(3, 1, 7) \) with respect to the plane.

1. **Identify the normal vector of the plane**: The equation of the plane is \( x - y + z = 3 \). The normal vector \( \mathbf{n} \) to this plane is given by the coefficients of \( x, y, z \), which is \( \mathbf{n} = (1, -1, 1) \).

2. **Find the foot of the perpendicular from point \( P \) to the plane**:
- Let the foot of the perpendicular be \( Q(x_0, y_0, z_0) \).
- The line through \( P \) in the direction of \( \mathbf{n} \) can be parameterized as:
\[ Q = P + t \mathbf{n} = (3 + t, 1 - t, 7 + t) \]
- Substitute \( Q \) into the plane equation:
\[ (3 + t) - (1 - t) + (7 + t) = 3 \]
Simplifying this:
\[ 3 + t - 1 + t + 7 + t = 3 \implies 3t + 9 = 3 \implies 3t = -6 \implies t = -2 \]
- Substitute \( t = -2 \) back into the parameterization to find \( Q \):
\[ Q = (3 - 2, 1 + 2, 7 - 2) = (1, 3, 5) \]

3. **Find the image point \( P' \)**: The image point \( P' \) is found by reflecting \( P \) across \( Q \):
\[ P' = Q + (Q - P) = (1, 3, 5) + ((1, 3, 5) - (3, 1, 7)) \]
\[ P' = (1, 3, 5) + (-2, 2, -2) = (-1, 5, 3) \]

### Step 2: Find the equation of the plane passing through \( P'(-1, 5, 3) \) and containing the line.

1. **Identify the direction ratios of the line**: The line is given by \( \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \). The direction ratios are \( (1, 2, 1) \).

2. **Find a point on the line**: A simple point on the line can be taken as \( (0, 0, 0) \).

3. **Find the normal vector to the plane**: The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of the direction vector of the line and the vector \( P'Q \):
- Let \( \mathbf{a} = (1, 2, 1) \) (direction ratios of the line) and \( \mathbf{b} = (-1 - 0, 5 - 0, 3 - 0) = (-1, 5, 3) \).
- The cross product \( \mathbf{n} = \mathbf{a} \times \mathbf{b} \):
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ -1 & 5 & 3 \end{vmatrix} = \mathbf{i}(2 \cdot 3 - 1 \cdot 5) - \mathbf{j}(1 \cdot 3 - 1 \cdot -1) + \mathbf{k}(1 \cdot 5 - 2 \cdot -1) \]
\[ = \mathbf{i}(6 - 5) - \mathbf{j}(3 + 1) + \mathbf{k}(5 + 2) = \mathbf{i}(1) - \mathbf{j}(4) + \mathbf{k}(7) = (1, -4, 7) \]

4. **Equation of the plane**: The equation of the plane can be written using the point-normal form:
\[ 1(x + 1) - 4(y - 5) + 7(z - 3) = 0 \]
Expanding this:
\[ x + 1 - 4y + 20 + 7z - 21 = 0 \implies x - 4y + 7z = 0 \]

### Final Answer:
The equation of the plane passing through point \( P'(-1, 5, 3) \) and containing the line is:
\[ x - 4y + 7z = 0 \]