Reflection of the line $\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z-3}{1}$ in the plane x + y + z = 7 is:

$\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$

Given line passes through (1, 2, 4) and this point also lies on the given plane.

Thus required line will be in the form of $\frac{x-1}{\ell}=\frac{y-2}{m}=\frac{z-4}{n}$

Any point on the given line is

(–r₁ + 1, 3r₁ + 2, r₁ + 4)

If r₁ = 1, this point becomes P = (0, 5, 5)

Let Q = (a, b, c) be the reflection of ‘P’ in the given plane, then

$\frac{a}{2} . 1+\frac{b+5}{2} . 1+\frac{5+c}{2} . 1=7$

i.e, a + b + c = 4,

and $\frac{a}{1}=\frac{b-5}{1}=\frac{c-5}{1}=\lambda$  (say)

$\Rightarrow a=\lambda, b=5+\lambda, c=5+\lambda$

$\Rightarrow 10+3 \lambda=4$

$\Rightarrow \lambda=-2$

Thus, Q = (–2, 3, 3)

Hence direction rations of reflected line are

–3, 1, –1

Thus it’s equation is

$\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$