Let the image of the point `P(2,-1,3)` in the plane `x+2y-z=0` be `Q`.Then the distance of the plane `3x+2y+z+29=0` from the point `Q` is

Options

A. `2sqrt(14)`
B. `(22sqrt(2))/(7)`
C. `3sqrt(14)`
D. `(24sqrt(2))/(7)`

Answer

To solve the problem step by step, we need to find the image of the point \( P(2, -1, 3) \) in the plane \( x + 2y - z = 0 \) and then calculate the distance from the point \( Q \) (the image of \( P \)) to the plane \( 3x + 2y + z + 29 = 0 \).

### Step 1: Find the normal vector of the first plane
The equation of the plane is given by:
\[ x + 2y - z = 0 \]
The normal vector \( \mathbf{n_1} \) to this plane can be derived from the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (1, 2, -1) \]

### Step 2: Find the distance from point \( P \) to the plane
The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane \( x + 2y - z = 0 \), we can rewrite it as:
\[ 1x + 2y - 1z + 0 = 0 \]
Thus, \( A = 1, B = 2, C = -1, D = 0 \).

Substituting the coordinates of point \( P(2, -1, 3) \):
\[ d = \frac{|1(2) + 2(-1) - 1(3) + 0|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|2 - 2 - 3|}{\sqrt{1 + 4 + 1}} = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2} \]

### Step 3: Find the coordinates of the image point \( Q \)
The coordinates of the image point \( Q \) can be found using the formula:
\[ Q = P - 2d \cdot \frac{\mathbf{n_1}}{|\mathbf{n_1}|} \]
Where \( d \) is the distance calculated in the previous step and \( |\mathbf{n_1}| \) is the magnitude of the normal vector:
\[ |\mathbf{n_1}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \]
Now substituting \( d \) and \( \mathbf{n_1} \):
\[ Q = (2, -1, 3) - 2 \cdot \frac{\sqrt{6}}{2} \cdot \frac{(1, 2, -1)}{\sqrt{6}} = (2, -1, 3) - (1, 2, -1) = (1, -3, 4) \]

### Step 4: Find the distance from point \( Q \) to the second plane
Now we need to find the distance from point \( Q(1, -3, 4) \) to the plane \( 3x + 2y + z + 29 = 0 \). The coefficients are \( A = 3, B = 2, C = 1, D = 29 \).

Using the distance formula:
\[ d = \frac{|3(1) + 2(-3) + 1(4) + 29|}{\sqrt{3^2 + 2^2 + 1^2}} = \frac{|3 - 6 + 4 + 29|}{\sqrt{9 + 4 + 1}} = \frac{|30|}{\sqrt{14}} = \frac{30}{\sqrt{14}} \]

### Final Answer
Thus, the distance from the point \( Q \) to the plane \( 3x + 2y + z + 29 = 0 \) is:
\[ \frac{30}{\sqrt{14}} \]