The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line

$3\left(2\right)+1(-1)+(-5)\left(1\right)=0$

Given line and given plane are parallel

$\therefore$ Image line is also parallel to the given line

Image of A (1,3,4) w.r.to given plane lies on the image line.

Equation of the normal to the plane is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$

Any point on the line $B=(2r+1,-r+3,r+4)$

If B is the image of A (1,3,4) then mid point of AB lies on the plane

Mid point $=\left(\frac{{\displaystyle 2r+2}}{{\displaystyle 2}},\frac{{\displaystyle -r+6}}{{\displaystyle 2}},\frac{{\displaystyle r+8}}{{\displaystyle 2}}\right)$

Mid point lies in the given plane $\Rightarrow 2\left(\frac{{\displaystyle 2r+2}}{{\displaystyle 2}}\right)-\left(\frac{{\displaystyle -r+6}}{{\displaystyle 2}}\right)+\frac{r+8}{2}+3=0$

$\Rightarrow r=-2, B(-3,5,2)$

Image lines is $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$