JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 83

The line's direction $(3,1,-5)$ is perpendicular to the plane's normal $(2,-1,1)$, since $6-1-5=0$. So the line is parallel to the plane and its image has the same direction.

Reflect $(1,3,4)$: $2-3+4+3=6$, so the image is $(1,3,4)-\frac{2\cdot6}{6}(2,-1,1)=(-3,5,2)$. The image line is
$$\frac{x+3}3=\frac{y-5}1=\frac{z-2}{-5}.$$