Let P be the image of the point (1, 2, 3) with respect to the plane x – y + z +1 = 0. Then the equation of the plane passing through P and containing the line $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-7}{1}$ is

$$\begin{gathered} \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{z-3}{1}=\frac{-2\times 3}{3} \\ \Rightarrow \mathrm{P}(-1,4,1)is the image of \left(1,2,3\right) w.r.t the planex-y+z+1=0 \\ \mathrm{Q}(3,1,7) is a point on the given line \\ drs of PQ=\left(4,-3,6\right) \end{gathered}$$

$normal to the plane=\overrightarrow{\mathrm{N}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 4& -3& 6\\ 1& 2& 1\end{array}\right|=-15\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+11\widehat{\mathrm{k}}$

$$\begin{gathered} Equation of the plane is 15\left(x+1\right)-2\left(y-4\right)-11\left(z-1\right)=0 \\ \text{ Required plane is: }15\mathrm{x}-2\mathrm{y}-11z+34=0 \end{gathered}$$