Finding the points in a plane that are equally far away from two points outside the plane. The plane is given: $2x-y+z-3=0$.
The point $P=(3,-4,2)$ is on one side of the plane and the point $Q=(-2,6,7)$ is on the other side of the plane. How do I determine the points in the plane which are equally far away from both $P$ and $Q$?
I tried to put PS = QS, but that didn't work. I really don't know what to do here.
Accidently left out that the points i need to find must be IN the plane $2x-y+z-3=0$ 

Let $S=(x,y,z)$ be point in the plane equidistance to the 2 given points $P$ and $R$
Through the distance formula (leaving off the square root) we have:
$(x-3)^2+(y--4)^2+(z-2)^2 = (x--2)^2 + (y-6)^2 + (z-7)^2$
Working this out, we end up with $x-2y+z=-6$ (Please check)
So now you have to find the line that intersects $x-2y+z=-6$ with $2x-y+z=3$ That line contains all the points we are looking for. Can you take it from here?
