We know that image $(x, y, z)$ of a point $\left(x_1, y_1, z_1\right)$ in a plane $a x+b y+c z+d=0$ is
$\begin{aligned}
\frac{x-x_1}{a} & =\frac{y-y_1}{b}=\frac{z-z_1}{c} \\
& =\frac{-2\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}
\end{aligned}$
Here, point is $(3,2,1)$ and plane is
$\begin{array}{rlrl}
& 2 x-y+3 z=7 \\
& \therefore & \frac{x-3}{2} & =\frac{y-2}{-1}=\frac{z-1}{3} \\
& =\frac{-2[2(3)-(2)+3(1)-7]}{2^2+1^2+3^2} \\
\Rightarrow & \frac{x-3}{2} & =\frac{y-2}{-1}=\frac{z-1}{3}=-2(0) \\
\Rightarrow & & x & =3, y=2, z=1
\end{array}$