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

The d.r.s. of the normal to the plane are \(1,-1,1\)
\(\therefore \quad\) The equation of line QM is
\(\begin{aligned}
& \frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\lambda(\text { say }) \\
& \Rightarrow x=\lambda+3, y=-\lambda+1, z=\lambda+7
\end{aligned}\)
Let \(M \equiv(\lambda+3,-\lambda+1, \lambda+7)\)
\(\therefore \quad\) Equation of plane becomes
\(\begin{aligned}
& 1(\lambda+3)-1(-\lambda+1)+1(\lambda+7)=3 \\
& \Rightarrow 3 \lambda+6=0 \Rightarrow \lambda=-2 \\
& \mathrm{M} \equiv(1,3,5)
\end{aligned}\)
Since \(M\) is the midpoint of PQ .
\(\therefore \quad \frac{3+\mathrm{a}}{2}=1, \frac{1+\mathrm{b}}{2}=3, \frac{7+\mathrm{c}}{2}=5\)
\(\Rightarrow a=-1, b=5, c=3\)
Equation of the plane passing through P and containing the given line is
\(\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y-5 & z-3 \\
1 & -5 & -3 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow x-4 y+7 z=0
\end{aligned}\)