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

Given line \(\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}\) passes through the point \((1,3,4)\)
Let the required line passes through the point (p, q, r)
Now, according to the given condition, we get Distance between point \((1,3,4)\) and the given plane \(=\) Distance between point \((p, q, r)\) and the given plane.
\(\therefore \quad\left|\frac{2(1)-(3)+(4)+3}{\sqrt{(2)^2+(-1)^2+(1)^2}}\right|=\left|\frac{2(\mathrm{p})-(\mathrm{q})+(\mathrm{r})+3}{\sqrt{(2)^2+(-1)^2+(1)^2}}\right|\)
\(\therefore \quad|2 \mathrm{p}-\mathrm{q}+\mathrm{r}+3|=6\)
Note that line given in option (D) passes through the point \((-3,5,2)\) and this point satisfies the condition given in equation (i).
\(\therefore \quad\) Option (D) is correct.