Orthogonal opposite edges of a tetrahedron

Place one vertex of a tetrahedron at the origin and denote the other vertex vectors by $a,b,c$. Orthogonality of two pairs of opposite edges of a tetrahedron forces $a\cdot b=a\cdot c=b\cdot c$. It follows that the third pair is orthogonal too. The three sums of squared lengths of opposite edges all equal $|a{|}^2+|b{|}^2+|c{|}^2-2a\cdot b$. This gives a dot product test for the special geometry without solving for the face planes.