the mirror image of point `(3,1,7)` with respect to the plane `x-y+z=3` is `P`. then equation plane which is passes through the point `P` and contains the line `x/1=y/2=z/1`.

Options

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

Explanation

To solve the problem step-by-step, we will first find the mirror image of the point \( (3, 1, 7) \) with respect to the plane \( x - y + z = 3 \). After that, we will find the equation of the plane that passes through the mirror image point \( P \) and contains the given line.

### Step 1: Find the mirror image of the point \( (3, 1, 7) \)

1. **Identify the point and the plane**: The point is \( A(3, 1, 7) \) and the plane is given by the equation \( x - y + z = 3 \).
2. **Find the normal vector of the plane**: The normal vector \( \vec{n} \) of the plane \( x - y + z = 3 \) is \( (1, -1, 1) \).
3. **Calculate the distance from the point to the plane**: The distance \( d \) from point \( A \) to the plane can be calculated using the formula:
\[ d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \]
where \( (a, b, c) \) are the coefficients of the plane equation and \( (x_1, y_1, z_1) \) are the coordinates of point \( A \).
Here, \( a = 1, b = -1, c = 1, d = 3 \), and \( (x_1, y_1, z_1) = (3, 1, 7) \):
\[ d = \frac{|1 \cdot 3 - 1 \cdot 1 + 1 \cdot 7 - 3|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{|3 - 1 + 7 - 3|}{\sqrt{3}} = \frac{|6|}{\sqrt{3}} = 2\sqrt{3} \]

4. **Find the foot of the perpendicular from point \( A \) to the plane**: The foot of the perpendicular \( F \) can be found using the formula:
\[ F = A - d \cdot \frac{\vec{n}}{|\vec{n}|} \]
Here, \( |\vec{n}| = \sqrt{3} \), so:
\[ F = (3, 1, 7) - 2\sqrt{3} \cdot \frac{(1, -1, 1)}{\sqrt{3}} = (3, 1, 7) - (2, -2, 2) = (1, 3, 5) \]

5. **Find the mirror image \( P \)**: The mirror image \( P \) can be calculated as:
\[ P = F + d \cdot \frac{\vec{n}}{|\vec{n}|} = (1, 3, 5) + (2, -2, 2) = (3, 1, 7) \]
Therefore, the coordinates of point \( P \) are:
\[ P = (1 - 2, 3 + 2, 5 - 2) = (-1, 5, 3) \]

### Step 2: Find the equation of the plane containing the line and point \( P \)

1. **Identify the line**: The line is given in symmetric form as:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \]
This can be represented in parametric form as:
\[ x = t, \quad y = 2t, \quad z = t \]
The direction ratios of the line are \( (1, 2, 1) \).

2. **Use point \( P \) and direction ratios to find the plane**: The plane can be defined using the normal vector which is perpendicular to the direction ratios of the line and passes through point \( P(-1, 5, 3) \). The normal vector can be found using the cross product of the direction vector of the line and the vector from point \( P \) to any point on the line.

3. **Choose a point on the line**: Let’s take \( t = 0 \) to get point \( (0, 0, 0) \) on the line.

4. **Vector from \( P \) to the point on the line**:
\[ \vec{v} = (0 - (-1), 0 - 5, 0 - 3) = (1, -5, -3) \]

5. **Find the normal vector using cross product**:
\[ \vec{n} = (1, 2, 1) \times (1, -5, -3) \]
The determinant gives:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 1 & -5 & -3 \end{vmatrix} = \hat{i}(2 \cdot (-3) - 1 \cdot (-5)) - \hat{j}(1 \cdot (-3) - 1 \cdot 1) + \hat{k}(1 \cdot (-5) - 2 \cdot 1) \]
\[ = \hat{i}(-6 + 5) - \hat{j}(-3 - 1) + \hat{k}(-5 - 2) = \hat{i}(-1) + \hat{j}(4) + \hat{k}(-7) \]
Thus, the normal vector is \( (-1, 4, -7) \).

6. **Equation of the plane**: The equation of the plane can be expressed as:
\[ -1(x + 1) + 4(y - 5) - 7(z - 3) = 0 \]
Simplifying this gives:
\[ -x - 1 + 4y - 20 - 7z + 21 = 0 \implies -x + 4y - 7z = 0 \]
or
\[ x - 4y + 7z = 0 \]

### Final Answer:
The equation of the plane is \( x - 4y + 7z = 0 \).