$A(1,-3,2),B(0,-4,5),C(5,0,-3)$ are points in $\mathbb{R}^3$.
I have found the vector equation of the line L through A and B.
$L=i-3j+2k+t(-i-j+3k)$
I have also found the foot of perpendicular from C to L, and hence determine the image of C under a reflection with respect to L.
Let $P$ be the foot of the perpendicular from $C$ to $L$.
$P = (3,-1,4)$
Since P is the midpoint between $C$ and $C'$, I have also found $C'$ which is $C'(1,-2,-5)$
However, now I am stuck at the below mentioned two questions, I do not know what to use and how to calculate.
iv) Find a vector equation of the image of the line through A and C under a reflection with respect to L.
v) Find the distance from C to L.
$\vec{BA}(1,1,-3),$ which gives an equation of $AB$: $$\vec{x}=(1,-3,2)+t(1,1,-3).$$ Let $P(1+t,-3+t,2-3t)$.
Thus, $CP(-4+t,-3+t,5-3t)$ and since $\vec{CP}\cdot\vec{BA}=0,$ we obtain: $$(1,1,-3)(-4+t,-3+t,5-3t)=0,$$ which gives $$t=2$$ and $$P\left(3,-1,-4\right).$$ Thus, since $P$ is a midpoint of $CC'$ we obtain: $$\frac{x_{C'}+5}{2}=3,$$ $$\frac{y_{C'}+0}{2}=-1$$ and $$\frac{z_{C'}-3}{2}=-4,$$ which gives $$C'\left(1,-2,-5\right).$$ Now, $$\vec{AC'}\left(0,1,-7\right)$$ and the needed equation it's: $$i-3j+2k+t(j-7k).$$ Also, we have $$CP=\sqrt{\left(-2\right)^2+\left(-1\right)^2+(-1)^2}=\sqrt{6}.$$