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

Options

A. `x-2y+3z=2`
B. `2x-4y+6z=7`
C. `x+2y-6z+2=0`
D. `x-2y+3z+2=0`

Answer

To solve the problem step-by-step, we will first find the image of the point \( P(3, 1, 7) \) with respect to the plane \( x - y + z = 3 \). Then, we will find the equation of the plane that passes through this image point and is parallel to the plane \( x - 2y + 3z = 7 \).

### Step 1: Find the image of the point \( P(3, 1, 7) \) with respect to the plane \( x - y + z = 3 \).

1. **Identify the coefficients of the plane equation**:
The equation of the plane can be written in the form \( ax + by + cz + d = 0 \).
Here, \( a = 1 \), \( b = -1 \), \( c = 1 \), and \( d = -3 \).

2. **Use the formula for the image of a point with respect to a plane**:
The image \( P'(x', y', z') \) of a point \( P(x_1, y_1, z_1) \) with respect to the plane is given by:
\[ P' = P + 2d_n \]
where \( d_n \) is the normal distance from the point to the plane.

The formula for the coordinates of the image is:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \]

3. **Calculate the distance**:
Plugging in the values:
\[ ax_1 + by_1 + cz_1 + d = 1(3) + (-1)(1) + 1(7) - 3 = 3 - 1 + 7 - 3 = 6 \]
The denominator is:
\[ a^2 + b^2 + c^2 = 1^2 + (-1)^2 + 1^2 = 1 + 1 + 1 = 3 \]
Therefore, we have:
\[ -\frac{6}{3} = -2 \]

4. **Set up the equations**:
Now, we can set up the equations:
\[ \frac{x - 3}{1} = \frac{y - 1}{-1} = \frac{z - 7}{1} = -2 \]

5. **Solve for \( x, y, z \)**:
From \( \frac{x - 3}{1} = -2 \):
\[ x - 3 = -2 \implies x = 1 \]
From \( \frac{y - 1}{-1} = -2 \):
\[ y - 1 = 2 \implies y = 3 \]
From \( \frac{z - 7}{1} = -2 \):
\[ z - 7 = -2 \implies z = 5 \]
Thus, the image point \( P' \) is \( (1, 3, 5) \).

### Step 2: Find the equation of the plane passing through \( P' \) and parallel to \( x - 2y + 3z = 7 \).

1. **Identify the normal vector of the given plane**:
The normal vector of the plane \( x - 2y + 3z = 7 \) is \( (1, -2, 3) \).

2. **Use the point-normal form of the plane**:
The equation of a plane with normal vector \( (a, b, c) \) passing through the point \( (x_1, y_1, z_1) \) is given by:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
Substituting \( (x_1, y_1, z_1) = (1, 3, 5) \) and \( (a, b, c) = (1, -2, 3) \):
\[ 1(x - 1) - 2(y - 3) + 3(z - 5) = 0 \]

3. **Expand and simplify**:
Expanding this gives:
\[ x - 1 - 2y + 6 + 3z - 15 = 0 \]
Simplifying:
\[ x - 2y + 3z - 10 = 0 \]

### Final Answer:
The equation of the plane passing through the point \( P' \) and parallel to the given plane is:
\[ x - 2y + 3z = 10 \]