We have,

Equation of line $PM$ is

$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda$

Any point on the line is

$M\equiv \left(\lambda +2,2\lambda -1,-\lambda +3\right)$

Point $M$ lies on the plane, so

$\left(\lambda +2\right)+2\left(2\lambda -1\right)-\left(3-\lambda \right)=0$

$\Rightarrow \lambda =\frac{1}{2}$

Point $M\equiv \left(\frac{5}{2},0,\frac{5}{2}\right)$

For Image $Q\left(\alpha ,\beta ,\gamma \right)$, we have

$\frac{\alpha +2}{2}=\frac{5}{2},\frac{\beta -1}{2}=0$

$\frac{\gamma +3}{2}=\frac{5}{2}$

So, $Q\equiv \left(3,1,2\right)$

Distance of the point $Q$ from the plane $3x+3y+z+29=0$ is

$d=\left|\frac{3\left(3\right)+2\left(1\right)+2+29}{\sqrt{3^2+2^2+1^2}}\right|$

$\Rightarrow d=\frac{42}{\sqrt{14}}=3\sqrt{14}$ units