$22 \cdot 12$ Let $O$ be the center of the base $\triangle ABC$ of a regular tetrahedron $P-ABC$. A moving plane through $O$ intersects the three lateral edges or their extensions of $P-ABC$ at points $Q, R, S$, respectively. Then the sum $\frac{1}{PQ} + \frac{1}{PR} + \frac{1}{PS}$
(A) has a maximum value but no minimum value.
(B) has a minimum value but no maximum value.
(C) has both a maximum value and a minimum value, and they are not equal.
(D) is a constant independent of the position of the plane $QRS$.
[Solution] The tetrahedron $P Q R S$ can be divided into three triangular pyramids with $O$ as the common vertex and $\triangle P Q R, \triangle P R S, \triangle P Q S$ as the bases.
Given that $\angle Q P R=\angle R P S=\angle S P Q=\alpha$.
Also, $O$ is the center of the base $\triangle A B C$ of the pyramid $P-A B C$, and the distance from $O$ to the three side faces is equal, which can be set as $d$. Therefore, the heights of the triangular pyramids $O-P Q R, O-P R S, O-P Q S$ are all $d$, so
$$
\begin{aligned}
V_{P Q R S S}= & V_{O-P Q R}+V_{O-P R S}+V_{O-P Q S} \\
= & \frac{1}{6} P Q \cdot P R \cdot \sin \alpha \cdot d+\frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot d \\
& +\frac{1}{6} P Q \cdot P S \cdot \sin \alpha \cdot d .
\end{aligned}
$$

On the other hand, the tetrahedron $P Q R S$ can also be seen as the triangular pyramid $Q-P R S$ with $Q$ as the vertex and $\triangle P R S$ as the base. Let the angle between $P A$ and the plane $P B C$ be $\theta$, then the height of this triangular pyramid is $P Q \sin \theta$.
Thus, we also have
$$
\begin{array}{l} 
V_{P Q R S}= V_{Q-R P S}=\frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot P Q \cdot \sin \theta . \\
\text { Therefore, } \quad P Q \cdot P R \cdot d+P R \cdot P S \cdot d+P S \cdot P Q \cdot d \\
= P R \cdot P S \cdot P Q \cdot \sin \theta,
\end{array}
$$

which means $\frac{1}{P S}+\frac{1}{P Q}+\frac{1}{P R}=\frac{\sin \theta}{d}$ is a constant.
Therefore, the answer is $(D)$.
Note If considering $Q, R, S$ on the extensions of the side edges, then $(B)$ is also correct.