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

Let $F$ be mid point of $\mathrm{AP}$

Line $\mathrm{AP} :\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\lambda$

$\Rightarrow F\left(3+\lambda , 1-\lambda , \lambda +7\right)$ lies in the plane

$\therefore 3+\lambda -\left(1-\lambda \right)+\lambda +7=3$

$3\lambda = -6 \Rightarrow \lambda =-2$

$\Rightarrow F\left(1,3,5\right)$

$\Rightarrow P\left(-1,5,3\right)$     (using mid point $F$)

So required plane (containing line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ & passing through $P\left(-1,5,3\right)$) is

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