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:

We have to find the mirror image (reflection) of the given line

$$\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$$

in the plane

$$2x-y+z+3=0.$$

First we convert the line into parametric form. Let the common ratio be $$t$$. Then

$$x=1+3t,\qquad y=3+1\cdot t,\qquad z=4-5t.$$

So every point on the line may be written as

$$P(t)\bigl(1+3t,\;3+t,\;4-5t\bigr).$$

The direction ratios (d.r.’s) of the line are the coefficients of $$t$$, that is

$$\vec d=\langle 3,\,1,\,-5\rangle.$$

Next we write the normal vector of the reflecting plane. For a plane $$ax+by+cz+d=0$$ the normal vector is $$\langle a,\,b,\,c\rangle.$$ Hence for the plane $$2x-y+z+3=0$$ we have

$$\vec n=\langle 2,\,-1,\,1\rangle.$$

Now we check whether the direction vector changes after reflection. For this we compute the dot-product

$$\vec d\cdot\vec n =3\cdot2+1\cdot(-1)+(-5)\cdot1 =6-1-5 =0.$$

Because $$\vec d\cdot\vec n=0,$$ the direction vector is already perpendicular to the normal, i.e. it lies entirely inside the plane. A vector that is parallel to the plane remains unchanged on reflection. Therefore the image line will have the same direction ratios $$\langle 3,\,1,\,-5\rangle.$$ So we only have to find one point on the image line.

Take the convenient point on the original line obtained by putting $$t=0$$:

$$P_0(1,\,3,\,4).$$

We now reflect this point in the plane. The standard formula for the image $$P'(x',y',z')$$ of a point $$P(x_0,y_0,z_0)$$ in the plane $$ax+by+cz+d=0$$ is

$$ x'=x_0-\frac{2a(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2},\\ y'=y_0-\frac{2b(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2},\\ z'=z_0-\frac{2c(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2}. $$

Here $$a=2,\;b=-1,\;c=1,\;d=3$$ and $$P_0(1,3,4).$$ First compute

$$S=ax_0+by_0+cz_0+d =2\cdot1+(-1)\cdot3+1\cdot4+3 =2-3+4+3 =6.$$

The denominator is

$$a^2+b^2+c^2 = 2^2+(-1)^2+1^2 = 4+1+1 = 6.$$

Substituting into the coordinate formulas:

$$ x' = 1-\frac{2\cdot2\cdot6}{6} =1-\frac{24}{6} =1-4 =-3,\\[4pt] y' = 3-\frac{2\cdot(-1)\cdot6}{6} =3+\frac{12}{6} =3+2 =5,\\[4pt] z' = 4-\frac{2\cdot1\cdot6}{6} =4-\frac{12}{6} =4-2 =2. $$

Thus the image point is

$$P'(-3,\,5,\,2).$$

The image line passes through this point and has the same d.r.’s $$\langle 3,\,1,\,-5\rangle.$$ Writing its symmetric form we get

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

This exactly matches Option C.

Hence, the correct answer is Option C.