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