Let p be a point outside the plane α where the triangle ABC lies. If PA, PB and PC are perpendicular to each other, then the photography of P in α is triangle A..

PA, PA, PC and △ABC (below) form a tetrahedron. If PA, PB and PC are perpendicular to each other, then the inclined planes are all right triangles.

The intersection point P is the height of the bottom of three inclined planes, and projected on the bottom surface △ABC is the height of three sides of △ABC, so the photography of P in α is the vertical center of triangle ABC.