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+7z=0`
D. `2x-y=0`

Answer

To solve the problem step by step, we will follow these steps:

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

1. **Identify the normal vector of the plane**: The equation of the plane can be rewritten as \(x - y + z - 3 = 0\). The normal vector \(N\) of the plane is given by the coefficients of \(x\), \(y\), and \(z\), which is \(N = (1, -1, 1)\).

2. **Find the point A**: The point A is given as \(A(3, 1, 7)\).

3. **Calculate the distance from point A to the plane**: The distance \(d\) from point A to the plane is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
where \(A = 1\), \(B = -1\), \(C = 1\), \(D = -3\), 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 A to the plane**: The foot of the perpendicular \(B\) can be found using the formula:
\[ B = A - d \cdot \frac{N}{\|N\|} \]
where \(\|N\| = \sqrt{3}\).
\[ B = (3, 1, 7) - 2\sqrt{3} \cdot \frac{(1, -1, 1)}{\sqrt{3}} = (3, 1, 7) - (2, -2, 2) = (1, 3, 5) \]

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

### Step 2: Find the equation of the plane passing through point P and containing the line given by \(\frac{x}{1} = \frac{y}{2} = \frac{z}{1}\).

1. **Identify the direction ratios of the line**: The line can be expressed 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 the direction ratios to find the equation of the plane**: The general equation of a plane can be expressed as:
\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane (in this case, point P) and \(A, B, C\) are the direction ratios of the line.

3. **Substituting P and the direction ratios**:
\[ 1(x + 1) + 2(y - 5) + 1(z - 3) = 0 \]
Expanding this gives:
\[ x + 1 + 2y - 10 + z - 3 = 0 \implies x + 2y + z - 12 = 0 \]

4. **Rearranging the equation**: The equation of the plane is:
\[ x + 2y + z = 12 \]

### Step 3: Finalize the equation of the plane.

The equation of the plane passing through point P and containing the line is:
\[ x + 2y + z = 12 \]