14.14. Two triangles - an equilateral one with side $a$ and an isosceles right one with legs equal to $b,$ are positioned in space such that their centers of mass coincide. Find the sum of the squares of the distances from all vertices of one to all vertices of the other.
14.14. If $M$ is the center of mass of triangle $ABC$, then $I_{M}=$ $=\left(AB^{2}+BC^{2}+AC^{2}\right) / 3$ (see problem 14.10, a), therefore for any point $X$ we have the equality $XA^{2}+XB^{2}+XC^{2}=I_{X}=$ $=3XM^{2}+I_{M}=3XM^{2}+\left(AB^{2}+BC^{2}+AC^{2}\right) / 3$. If $ABC$ is a given right triangle, $A_{1}B_{1}C_{1}$ is a given equilateral triangle, and $M$ is their common center of mass, then $A_{1}A^{2}$ $+A_{1}B^{2}+A_{1}C^{2}=3A_{1}M^{2}+4b^{2} / 3=a^{2}+4b^{2} / 3$. Writing similar equalities for points $B_{1}$ and $C_{1}$ and adding them, we get that the desired sum of squares is $3c^{2}+4b^{2}$.