The reflection of the point $(2, -1, 3)$ in the plane $3x-2y-z=9$is:

Let $P$ be the point $(2, -1, 3)$ and $Q$ be its reflection in the given plane. 

Then, $PQ$ is perpendicular to the given plane

Hence, d.r.'s of $PQ$ are $3, -2, 1$ and consequently, equations of 

 $\mathrm{PQ}$ are $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-3}{-1}$

Any point on this line is $(3r+2,-2r-1,-r+3)$

Let this point be $Q.$ Then midpoint of $PQ$

$\begin{array}{l}=\left(\frac{3r+2+2}{2},\frac{-2r-1-1}{2},\frac{-r+3+3}{2}\right)\\ =\left(\frac{3r+4}{2},-r-1,\frac{-r+6}{2}\right)\end{array}$

This point lies in given plane i.e. 

$\begin{array}{l}3\left(\frac{3r+4}{2}\right)-2(-r-1)-\left(\frac{-r+6}{2}\right)=9\\ \Rightarrow 9r+12+4r+4+r-6=9\\ \Rightarrow 14r=8\Rightarrow r=\frac{4}{7}\end{array}$

Hence, the required point $Q$ is

$\left(3\left(\frac{4}{7}\right)+2,-2\left(\frac{4}{7}\right)-1,\frac{-4}{7}+3\right)=\left(\frac{26}{7},\frac{-15}{7},\frac{17}{7}\right)$