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 $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$

Let image (x, y, z) of

$\therefore \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}-7}{1}=-2\left(\frac{3-1+7-3}{1^2+1^2+1^2}\right)=-4$

$\Rightarrow \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(-1, 5, 3)$ 

Equation of plane passing through P(-1,5,3)

Also  containing the  line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$ is

$$\begin{gathered} \left|\begin{array}{ccc}x+1& y-5 & z-3\\ 1& -5& -3\\ 1& 2& 1\end{array}\right|=0 \\ \Rightarrow \left(x+1\right)\left(-5+6\right)-\left(y-5\right)4+\left(z-3\right)7=0 \\ \Rightarrow x-4y+7z=0 \end{gathered}$$