13. (USA 6) ${ }^{\mathrm{IMO} 2}$ Given any point $P$ in the interior of a sphere with radius $R$, 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 \( O M^{2} + O P^{2} = O N^{2} + O Q^{2} \). 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 \( \left(O, \sqrt{3 R^{2} - O P^{2}}\right) \). 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 \( \gamma \); 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 \( P, A, B, C, X, Q \).