Reflection of the line $\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z-4}{1}$ in the plane $x+y+z=7$ is

Given line $\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z-4}{1}$ is passes through $P(1,2,4)$.

To find the reflection of line we need one more point on the line clearly $M(0,5,5)$ also lie on line.

Let $N(\alpha ,\beta ,\gamma )$ be the reflection $M$ in the $x+y+z=7$

$\therefore \frac{\alpha }{2}+\frac{\beta +5}{2}+\frac{\gamma +5}{2}=7$

$\Rightarrow \alpha +\beta +\gamma =4$

Also, $MN$ is perpendicular, i.e. parallel to the normal of the plane

$\therefore \frac{\alpha }{1}=\frac{\beta -5}{1}=\frac{\gamma -5}{1}=\lambda$

$\Rightarrow \alpha =\lambda ,\beta =\lambda +5,\gamma =\lambda +5$

$\therefore \lambda +\lambda +5+\lambda +5=4$

$\Rightarrow \lambda =-2$

$\therefore N=(-2,3,5)$

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