I am given the following problem:
Given $A = (2,1,3)$ and $B=(4,-1,1)$ and the plane $\alpha \{ \ 2x-y+2z=3$ find the parametric equations of the line $r$ contained in $\alpha$ such that, for every point in $r$, the distance between $r$ and $A$ is the same as the distance between $r$ and $B$
I am struggling to find the equations that will solve this. I know that the distance between a line and a point is
$$ \delta = \frac{\vert \vec{AP} \wedge \vec{u} \vert}{\Vert \vec{u} \Vert} $$
if $P$ is a point contained on the line and $\vec{u}$ is the vector that gives the direction of $r$. Also, the normal vector of the plane is
$$ \vec{n} = (2,-1,2)\\ $$
If I say that $\vec{u} = (a,b,c)$ I will have a LOT of variables and a few equations to solve (it won't get me nowhere.
but I honestly cannot go further than this. What piece of information am I missing?
Hint:
It is the intersection of the plane $\alpha$ and perpendicular bisector plane of segment $[AB]$. The latter is the plane through the midpoint $I$ of $[AB]$, normal to vector $\overrightarrow{AB}$.
For this perpendicular bisector, you should find the cartesian equation $x-y-z=1$.
From there you can easily find a parametric representation of the intersection line.