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 

$$\begin{gathered} Let M= foot of the perpendicular drawn from P to the plane. \\ \text{ eq. of line PM }\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda \\ \text{ any point on line }=(\lambda +2,2\lambda -1,-\lambda +3) \\ \text{ for point ' }M\text{ ' }(\lambda +2)+2(2\lambda -1)-(3-\lambda )=0 \\ \lambda =\frac{1}{2} \\ foot=M=\left(\frac{1}{2}+2,2\times \frac{1}{2}-1,\frac{-1}{2}+3\right)=\left(\frac{5}{2},0,\frac{5}{2}\right) \end{gathered}$$
Image of P(2, -1, 3) in the plane x + 2y - z = 0 is (3, 1, 2)
Distance of the plane 3x + 2y + z + 29 = 0 from the point Q is $=\left|\frac{9+2+2+29}{\sqrt{14}}\right|=3\sqrt{14}$