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

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

Clearly, given line is parallel to the given plane. So its image will also be parallel to the plane and so its direction ratios are proportional to 3, 1, -5. Given line passes through (1, 3, 4). Let $(x_1, y_1, z_1)$ be its image in the given plane.

Then ,  $x_1, y_1, z_1$ are given by 

$\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}= -2$

$⇒\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}= -2$

$⇒ x_1 = - 3, y_1 = 5, z_1 =  2$

So, image of (1, 3, 4) in the given plane is (-3, 5, 2).

Hence, the equations of the required line are

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