6. 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, which are not equal
(D) is a constant independent of the plane $QPS$
(D)
6.【Analysis and Solution】As shown in the figure, 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 $\angle Q P R = \angle P R S$
$$
= \angle P Q S = \alpha.
$$

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 $O-P Q R$, $O-P R S$, and $O-P Q S$ are all $d$, so we have
$$
\begin{aligned}
V_{P Q R S} = & V_{o-P Q R} + V_{o-P R S} + V_{o-P Q S} \\
= & \frac{1}{6} P Q \cdot P R \cdot \sin \alpha \cdot d + \frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot d \\
& + \frac{1}{6} P Q \cdot P S \cdot \sin \alpha \cdot d.
\end{aligned}
$$

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. If the angle between $P A$ and the plane $P B C$ is $\theta$, then the height of this triangular pyramid is $P Q \sin \theta$.
Thus, we also have $V_{P Q R S} = V_{Q-R P S} = \frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot P Q \cdot \sin \theta$,
$$
\begin{array}{l}
\therefore P Q \cdot P R \cdot d + P R \cdot P S \cdot d + P S \cdot P Q \cdot d \\
\quad P R \cdot P S \cdot P Q \cdot \sin \theta.
\end{array}
$$

Therefore, $\frac{1}{P S} + \frac{1}{P Q} + \frac{1}{P R} = \frac{\sin \theta}{d}$ is a constant, so the answer is (D).