13.3.3 * In a sphere, there is a fixed point $P$, and there are three moving points $A$, $B$, and $C$ on the sphere's surface. $\angle B P A=\angle C P A=\angle C P B=90^{\circ}$. A parallelepiped is formed with $P A$, $P B$, and $P C$ as edges, and point $Q$ is the vertex of the parallelepiped that is diagonally opposite to $P$. When $A$, $B$, and $C$ move on the sphere's surface, find the trajectory of point $Q$.
Parse the figure to establish a coordinate system. Let the center of the sphere be $(a, b, c)$, and the radius be $R$, then the equation of the sphere is $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}$ $=R^{2}$. From this, the coordinates of points $A$, $B$, and $C$ are
$$
A\left(a+\sqrt{R^{2}-b^{2}-c^{2}}, 0,0\right), B\left(0, b+\sqrt{R^{2}-a^{2}-c^{2}}, 0\right), C\left(0,0, c+\sqrt{R^{2}-a^{2}-b^{2}}\right) \text {. }
$$

Therefore, the coordinates of point $Q$ are $\left(a+\sqrt{R^{2}-b^{2}-c^{2}}, b+\sqrt{R^{2}-a^{2}-c^{2}}, c+\right.$ $\left.\sqrt{R^{2}-a^{2}-b^{2}}\right)$, so $|O Q|=\sqrt{3 R^{2}-2\left(a^{2}+b^{2}+c^{2}\right)}$. Since $R$ and $a^{2}+b^{2}+c^{2}=$ $O P^{2}$ are constants, $O Q$ is a constant. This indicates that the trajectory of point $Q$ is a spherical surface with $O$ as the center and $\sqrt{3 R^{2}-2 O P^{2}}$ as the radius.