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 through P and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$, is

$x - 4y + 7z = 0 $

We know that the image of the point $(x_1, y_1, z_1)$ in the plane ax + by + cz + d = 0 is given by

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$

Let P(x, y z) be the image of the point (3, 1, 7) in the plane x - y + z = 3. Then,

$\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\frac{-2(3-1+7-3)}{1^2+(-1)^2+1^2}$

$⇒ x = -1, y = 5 , z = 3 $

So, the image of  the given point is P(-1, 5, 3).

The equation of a plane passing through P(-1, 5, 3) is

$a(x+1) + b (y - 5) + c(z-3)= 0 $   ..............(i)

This will contain the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$, if

$ a - 5b - 3c = 0 $

and, $ a + 2b + c = 0 $

$∴ \frac{a}{-5+6}=\frac{b}{-3-1}=\frac{c}{2+5}$ or $\frac{a}{1}=\frac{b}{-4}=\frac{c}{7}$

Substituting the values of a, b, c in (i), we obtain x - 4y + 7z = 0 as the required plane.