3. In tetrahedron $ABCD$, $AB=3, CD=2$, the distance between line $AB$ and line $CD$ is 2, and the angle between them is $\frac{\pi}{6}$. Then the volume of tetrahedron $ABCD$ is $\qquad$
3. $\frac{1}{2}$.

Solution: As shown in the figure, draw $A F / / B C$ and $A F=B C$, draw $D E / / B C$ and $D E=B C$, connect $B E$, $D F$, and $B E$, to form a triangular prism $A E B-D F C$.
$$
\begin{array}{l}
\text { Then } V_{A B C D}=\frac{1}{3} V_{\text {prism }}=\frac{1}{3} \times \frac{1}{2} F C \cdot C D \sin \frac{\pi}{6}=\frac{1}{12} \\
\times 3 \times 2=\frac{1}{2} .
\end{array}
$$