3. Let the tetrahedron $ABCD$ be a regular tetrahedron, and $E$ be the midpoint of edge $BC$. Then the distance $d$ between the skew lines $AE$ and $CD$ is $\qquad$ .
3. $\frac{4 \sqrt{22}}{11}$.

As shown in Figure 2, let $F$ be the midpoint of side $BD$, and connect $AF$, $EF$, and $CF$. Then $d$ is the distance from point $C$ to the plane $AEF$.

Given that the edge length of the regular tetrahedron $a=4$, its volume is
$$
V=\frac{\sqrt{2}}{12} a^{3}=\frac{16 \sqrt{2}}{3} .
$$

Therefore, $d S_{\triangle AEF}=3 V_{1-CEF}=3 \times \frac{1}{4} V=4 \sqrt{2}$.
$$
\begin{array}{l}
\text { Also, } S_{\triangle AEF}=\frac{1}{2} EF \cdot h_{A}=h_{A} \\
=\sqrt{AE^{2}-\left(\frac{EF}{2}\right)^{2}} \\
=\sqrt{\left(\frac{\sqrt{3}}{2} \times 4\right)^{2}-1^{2}}=\sqrt{11},
\end{array}
$$

Thus, $d=4 \sqrt{\frac{2}{11}}=\frac{4 \sqrt{22}}{11}$.