10. Let $P$ be a fixed point inside a sphere, and draw three mutually perpendicular rays from $P$ intersecting the sphere at points $A$, $B$, $C$. Point $Q$ is the vertex of the rectangular parallelepiped formed by $P A$, $P B$, $P C$ that is diagonally opposite to $P$. Find the locus of all points $Q$ corresponding to all sets of three such rays drawn from $P$.
10. Solution Let the center of the sphere be $O$, and the radius of the sphere be $R$. Take $O$ as the origin, and $OP$ as the $x$-axis, to establish a rectangular coordinate system.

Let $OP=p$, then the coordinates of point $P$ are $(p, 0,0)$, and the equation of the sphere is $x^{2}+y^{2}+z^{2}=R^{2}$.
Let $Q(x, y, z), A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right), C\left(x_{3}, y_{3}, z_{3}\right)$,
then $x_{i}^{2}+y_{i}^{2}+z_{i}^{2}=R^{2}(i=1,2,3)$.
Since $A, B, C, P, Q$ are vertices of a rectangular box, and $PQ$ is the diagonal of the box,
we have $PQ^{2}=PA^{2}+PB^{2}+PC^{2}$,
i.e., $(x-p)^{2}+y^{2}+z^{2}=\left(x_{1}-p\right)^{2}+y_{1}^{2}+z_{1}^{2}+\left(x_{2}-p\right)^{2}+y_{2}^{2}+z_{2}^{2}+\left(x_{3}-p\right)^{2}+y_{3}^{2}+z_{3}^{2}$.
i.e.,
$$
(x-p)^{2}+y^{2}+z^{2}=3 R^{2}+3 p^{2}-2 p\left(x_{1}+x_{2}+x_{3}\right) .
$$

The coordinates of the midpoint of diagonal $PQ$ are $\left(\frac{x+p}{2}, \frac{y}{2}, \frac{z}{2}\right)$.
The coordinates of the vertex $A^{\prime}$, which is diagonally opposite to point $B$ in the rectangular box, are
$$
\overrightarrow{OA}+\overrightarrow{PC}=\left(x_{1}, y_{1}, z_{1}\right)+\left(x_{3}-p, y_{3}, z_{3}\right)=\left(x_{1}+x_{3}-p, y_{1}+y_{3}, z_{1}+z_{3}\right) .
$$

Thus, the coordinates of the midpoint of diagonal $A^{\prime}B$ are
$$
\left(\frac{x_{1}+x_{2}+x_{3}-p}{2}, \frac{y_{1}+y_{2}+y_{3}}{2}, \frac{z_{1}+z_{2}+z_{3}}{2}\right)
$$

Since the midpoint of diagonal $PQ$ coincides with the midpoint of diagonal $A^{\prime}B$, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}-p=x+p \\
y_{1}+y_{2}+y_{3}=y \\
z_{1}+z_{2}+z_{3}=z
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=2 p+x \\
y_{1}+y_{2}+y_{3}=y \\
z_{1}+z_{2}+z_{3}=z
\end{array}\right.\right.
$$

Substituting (2) into (1) to eliminate $x_{i}, y_{i}, z_{i}(i=1,2,3)$, we get $x^{2}+y^{2}+z^{2}=3 R^{2}-2 p^{2}$.
Therefore, the locus of point $Q$ is a sphere with center $O$ and radius $\sqrt{3 R^{2}-2 p^{2}}$.