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

The set of points at an equal distance from two given points is a plane which is perpendicular to the line joining the two points. The difference between the two points is a vector parallel to such line. Writing the planes perpendicular to a given vector is immediate. Finally force the plane to pass through the midpoint of the two given points.
