3. Given a parallelepiped $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with volume $V$, and points $P$, $Q$, $R$ are the midpoints of edges $A^{\prime} B^{\prime}$, $C C^{\prime}$, $C D$ respectively. Then the volume of the tetrahedron $A-P Q R$ is ( ).
(A) $\frac{V}{12}$
(B) $\frac{V}{16}$
(C) $\frac{V}{24}$
(D) $\frac{V}{28}$
3. C.

Assume the section $P Q A$ intersects the plane $A B C D$ at $A E$ (point $E$ is on $C D$). Then, by the planes $A B B^{\prime} A^{\prime} \parallel$ plane $D C C^{\prime} D^{\prime}$, we have $Q E \parallel P A$. Therefore, draw a line through point $Q$ parallel to $P A$ intersecting $C D$ at point $E$, and extend $A E$ and $P Q$ to intersect at point $N$ in the plane $P Q A$.
Since $P A \perp C^{\prime} R$, it follows that $Q E \parallel C^{\{\prime} R$.
Given that $Q$ is the midpoint of $C C^{\prime}$, then $E$ is the midpoint of $R C$.
Thus, $Q E \perp \frac{C R}{2} \perp \frac{P A}{2}$.
Therefore, $E$ is the midpoint of $A N$.
Hence, $S_{\triangle A P Q}=S_{\triangle M Q N}, V_{R-A P Q}=V_{R-A Q N}$.
Connecting $R N$, we have $S_{\triangle A R E}=S_{\triangle E R V}$.
Since $Q$ is the midpoint of $C C^{\prime}$, let the height of the parallelepiped be $h$, then the height of the pyramid $Q-A R N$ is $\frac{h}{2}$.
But $S_{\triangle R A N}=2 S_{\triangle A R E}=\frac{1}{4} S_{\square A B C D}$, so,
$$
V_{R-A Q N}=V_{Q-A R N}=\frac{1}{3} \times \frac{1}{4} S_{\square A B C D} \frac{h}{2}=\frac{V}{24} .
$$