Find a point equidistant from two points and a line

Question

Given a line: $x-\frac{2}{2} = y-\frac{1}{6} = z+\frac{2}{2}$ and two points $P_1(1,1,0)$ and $P_2(0,1,-1)$, identify the point $V$ which resides on the line and is equidistant from $P_1$, and $P_2$.

Answer 1

The locus of all the point $P(x,y)$ such that $PP_1=PP_2$ is the plane $$(x-1)^2+(y-1)^2+z^2=x^2+(y-1)^2+(z+1)^2\to x+z=0$$ Find the intersection between the line and the plane solving the system $$\begin{cases} \frac{x-2}{2} =\frac{y-1}{6}\\ \frac{x-2}{2} = \frac{z+2}{2}\\ x+z=0\\ \end{cases} $$ Solution should be $V(0,-5,0)$