3. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, points $E, F$ are the centers of the planes $B B_{1} C_{1} C$ and $A B C D$, respectively. Then the distance between the skew lines $E F$ and $A_{1} C_{1}$ is $\qquad$ .
3. $\frac{\sqrt{3}}{3}$.

As shown in Figure 2, connect $A_{1} C, A B_{1}$. Since $A_{1} C_{1} / / A C$, the distance from line $E F$ to $A_{1} C_{1}$ is equal to the distance from line $A_{1} C_{1}$ to the plane $A B_{1} C$, which is also equal to the distance from point $A_{1}$ to the plane $A B_{1} C$. Let this distance be $h$. Then
$$
V_{A_{1}-A B_{1} C}=\frac{1}{3} h \cdot S_{\triangle A B_{1} C}=\frac{\sqrt{3}}{6} h .
$$

Also,
$$
V_{A_{1}-A B_{1} C}=V_{C-A A_{1} B}=\frac{1}{3} \cdot 1 \cdot \frac{1}{2}=\frac{1}{6} .
$$

Thus, $\frac{\sqrt{3}}{6} h=\frac{1}{6}$, solving for $h$ gives $h=\frac{\sqrt{3}}{3}$.