Given a triangle ABC and a straight line L.

Find the point P on L such that

** PA**^{2}+PB^{2}+PC^{2}
is the smallest.

(In reply to

re: one messy method by Steven Lord)

Ah! The math reduces even more simply:

If a,b,c are the location of points on L that are the the dropped perpendiculars of A, B, C, then P is located at the average of these three points.