dius $R$, three mutually perpendicular segments $P A, P B, P C$ are drawn terminating on the sphere and having one common vertex in $P$. Consider the rectangular parallelepiped of which $P A, P B, P C$ are coterminal
edges. Find the locus of the point $Q$ that is diagonally opposite $P$ in the parallelepiped when $P$ and the sphere are fixed.
13. Lemma. If $M N P Q$ is a rectangle and $O$ any point in space, then $O M^{2}+$
$$
O P^{2}=O N^{2}+O Q^{2} \text {. }
$$

Proof. Let $O_{1}$ be the projection of $O$ onto $M N P Q$, and $m, n, p, q$ denote the distances of $O_{1}$ from $M N, N P, P Q, Q M$, respectively. Then $O M^{2}=O O_{1}^{2}+q^{2}+m^{2}, O N^{2}=O O_{1}^{2}+m^{2}+n^{2}, O P^{2}=O O_{1}^{2}+n^{2}+p^{2}$, $O Q^{2}=O O_{1}^{2}+p^{2}+q^{2}$, and the lemma follows immediately.

Now we return to the problem. Let $O$ be the center of the given sphere $S$, and $X$ the point opposite $P$ in the face of the parallelepiped through $P, A, B$. By the lemma, we have $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$ and $O P^{2}+$ $O X^{2}=O A^{2}+O B^{2}$. Hence $2 O P^{2}+O Q^{2}=O A^{2}+O B^{2}+O C^{2}=3 R^{2}$, i.e. $O Q=\sqrt{3 R^{2}-O P^{2}}>R$.

We claim that the locus of $Q$ is the whole sphere $\left(O, \sqrt{3 K^{2}-O P^{2}}\right)$. Choose any point $Q$ on this sphere. Since $O Q>R>O P$, the sphere
with diameter $P Q$ intersects $S$ on a circle. Let $C$ be an arbitrary point on this circle, and $X$ the point opposite $C$ in the rectangle $P C Q X$. By the lemma, $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$, hence $O X^{2}=2 R^{2}-O P^{2}>R^{2}$. The plane passing through $P$ and perpendicular to $P C$ intersects $S$ in a circle 7 ; both $P, X$ belong to this plane, $P$ being inside and $X$ outside the circle, so that the circle with diameter $P X$ intersects $\gamma$ at some point $B$. Finally, we choose $A$ to be the point opposite $B$ in the rectangle $P B X A$ : we deduce that $O A^{2}+O B^{2}=O P^{2}+O X^{2}$, and consequently $A \in S$. By the construction, there is a rectangular parallelepiped through $P, A, B, C, X, Q$