Example 10 (1995 National High School League Question) Let $O$ be the center of the base $\triangle ABC$ of the 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 $\frac{1}{PQ}+\frac{1}{PR}+\frac{1}{PS}(\quad)$.
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 the maximum value is not equal to the minimum value
D. is a constant independent of the position of the plane $QRS$
Solution: Choose D. Reason: As shown in Figure 7-9, 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, respectively. Given that $\angle Q P R=\angle R P S=\angle Q P S=\alpha$.

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 $O-P Q R, O-P Q S$ are both $d$, so we have $V_{\text {PQRS }}=V_{\text {OrRoR }}+V_{\text {OPRS }}+V_{\text {OPQS }}=\frac{1}{6} P Q \cdot P R \cdot \sin \alpha \cdot d+\frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot d+\frac{1}{6} P Q \cdot P S \cdot \sin \alpha \cdot d$.

On the other hand, the tetrahedron $P Q R S$ can also be seen as a triangular pyramid $Q-P R S$ with $Q$ as the vertex and $\triangle P R S$ as the base. If the angle between $P A$ and the plane $P B C$ is $\theta$, then the height of this triangular pyramid is $P Q \cdot \sin \theta$. Therefore, we also have $V_{\text {PQRS }}=V_{Q P R S}=\frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot P Q \cdot \sin \theta$, thus $P Q \cdot P R \cdot d+P R \cdot P S \cdot d+P S \cdot P Q \cdot d=P R \cdot P S \cdot P Q \cdot \sin \theta$, i.e., $\frac{1}{P S}+\frac{1}{P Q}+\frac{1}{P R}=\frac{\sin \theta}{d}$, which is a constant.