14.13. Given a triangle $ABC$. Find the geometric locus of points $X$ in space such that $X A^{2}+$ $+X B^{2}=X C^{2}$.
14.13. Place masses +1 at vertices $A$ and $B$, and a mass -1 at vertex $C$. The center of mass $M$ of this system of points is the diagonal of the parallelogram $A C B M$. According to the condition, $I_{X}=X A^{2}+X B^{2}-X C^{2}=0$, and since $I_{X}=(1+1-1) M X^{2}+I_{M I}$ (problem 14.9), then $M X^{2}=-I_{M}=a^{2}+b^{2}-c^{2}$, where $a, b$ and $c$ are the lengths of the sides of triangle $A B C$ (problem 14.10, b). Thus, if $\angle C<90^{\circ}$, the desired locus is a sphere of radius $\sqrt{a^{2}+b^{2}-c}$ centered at $M$.