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, and the maximum value is not equal to the minimum value;
(D) is a constant independent of the position of the plane $QRS$.


6. (D)

Tetrahedron $P Q R S$ can be divided into three triangular pyramids with a common vertex at point $O$, and bases $\triangle P Q R$, $\triangle P R S$, and $\triangle P Q S$ respectively.

Given that $\angle Q P R = \angle R P S = \angle P Q S = \alpha$.

Also, $O$ is the center of the base $\triangle A B C$ of the tetrahedron $P-A B C$, and the distances from point $O$ to the three side faces are equal, denoted as $d$. Therefore, the heights of the triangular pyramids $O-P Q R$, $O-P R S$, and $O-P Q S$ are all $d$. Thus, 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 viewed as a triangular pyramid with vertex $Q$ and base $\triangle P R S$, denoted as $Q-P R S$. 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 $V_{P Q R S} = V_{Q R P S} = \frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot P Q \cdot \sin \theta$,
$$
\therefore P Q \cdot P R \cdot d + P R \cdot P S \cdot d + P S \cdot P Q \cdot d = P R \cdot P S \cdot P Q \cdot \sin \theta \text{. }
$$

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