6. Let $O$ be the center of the base $\triangle A B C$ of the regular tetrahedron $P-A B C$. A moving plane through $O$ intersects the edges or their extensions of $P-A B C$ at points $Q, R, S$. The expression $\frac{1}{P Q}+\frac{1}{P R}+\frac{1}{P S}$
(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 $Q R S$
6. (D).

Tetrahedron $P Q R S$ can be divided into three pyramids with a common vertex $O$ and bases $\triangle P Q R, \triangle P R S, \triangle P Q S$.
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 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 pyramids $O-P Q R$, $O-P R S$, and $O-P Q S$ are all $d$. Thus, $V_{P Q R S}=V_{O-P Q R}+V_{\text {aPRS }}+V_{Q-P Q S}=\frac{1}{6} P Q \cdot P R \cdot \sin \alpha \cdot d+\frac{1}{6} P R \cdot P S \cdot \sin \alpha \cdot d+\frac{1}{6} P Q \cdot P S \cdot \sin \alpha \cdot d$.

On the other hand, tetrahedron $P Q R S$ can also be seen as a pyramid $Q-P R S$ with vertex $Q$ and base $\triangle P R S$. Let the angle between $P A$ and the plane $P B C$ be $\theta$. Then the height of this 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, $P Q \cdot P R \cdot d+P R \cdot P S \cdot d+P S \cdot P Q \cdot d=P R \cdot P S \cdot P Q \cdot \sin \theta$, which simplifies to $\frac{1}{P S}+\frac{1}{P Q}+\frac{1}{P R}=\frac{\sin \theta}{d}$, a constant.

Note: If considering $Q, R, S$ on the extensions of the side edges, then (B) is also correct.