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 $\frac{1}{PQ} + \frac{1}{PR} + \frac{1}{PS}$ (
(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 $QRS$
)
Consider using the transformation of equal volume. As shown in the figure, let the three dihedral angles at vertex $P$ be $\alpha$, and the angle between a lateral edge and the opposite face be $\theta$. If we consider the tetrahedron $P Q R S$ as a triangular pyramid with $Q$ as the vertex and $\triangle P R S$ as the base, we easily get
$$
V_{P-Q R S}=V_{Q-P R S}=\frac{1}{6} P R \cdot P S \sin \alpha \cdot P Q \sin \theta.
$$

The tetrahedron $P Q R S$ can also be divided into three triangular pyramids with $O$ as the common vertex and $\triangle P Q R$, $\triangle P R S$, $\triangle P S Q$ as the bases, respectively. Since $O$ is the center of the base of the regular triangular pyramid $P-A B C$, the distance from $O$ to the three faces is equal, denoted as $d$. Therefore,
$$
\begin{array}{l}
V_{P-Q R S}=V_{O-P Q R}+V_{O-P R S}+V_{O-P S Q}=\frac{1}{6}(P Q \cdot P R+P R \cdot P S+P Q \cdot P S) \cdot d \cdot \sin \alpha. \\
\text { Thus, }(P Q \cdot P R+P R \cdot P S+P Q \cdot P S) d=P R \cdot P Q \cdot P S \cdot \sin \theta, \text { so } \frac{1}{P S}+\frac{1}{P Q}+ \\
\frac{1}{P R}=\frac{\sin \theta}{d} \text { is a constant. Therefore, the answer is D. } \\
\end{array}
$$