124.6 ** In a sphere, there is a fixed point $P$, and there are three moving points $A$, $B$, $C$ on the sphere's surface. $\angle B P A = \angle C P A = \angle C P B = 90^{\circ}$. A parallelepiped is formed with $P A$, $P B$, $P C$ as edges, and point $Q$ is the vertex of the parallelepiped that is diagonally opposite to point $P$. When $A$, $B$, $C$ move on the sphere's surface, find the trajectory of point $Q$.
Parse: Establish a spatial rectangular coordinate system $P-ABC$, let the center of the sphere be $O(a, b, c)$, and the radius be $R$. Then $O P^{2}=a^{2}+b^{2}+c^{2}$ is a constant, and the equation of the sphere is $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}-R^{2}$. From this, we get the coordinates of points $A$, $B$, and $C$ as $A\left(a+\sqrt{R^{2}-b^{2}-c^{2}}, 0,0\right)$, $B(0, b+$ $\left.\sqrt{R^{2}-a^{2}-c^{2}}, 0\right)$, and $C\left(0,0, c+\sqrt{R^{2}-a^{2}-b^{2}}\right)$. Therefore, the coordinates of point $Q$ are
$$
Q\left(a+\sqrt{R^{2}-b^{2}-c^{2}}, b+\sqrt{R^{2}-a^{2}-c^{2}}, c+\sqrt{R^{2}-a^{2}-b^{2}}\right),
$$

Thus, $|O Q|=\sqrt{3 R^{2}-2\left(a^{2}+b^{2}+c^{2}\right)}=\sqrt{3 R^{2}-2 \cdot O P^{2}}$. This shows that the trajectory of point $Q$ is a spherical surface with $O$ as the center and $\sqrt{3 R^{2}-2 \cdot O P^{2}}$ as the radius.