3. Given a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, the distance between the skew lines $A C_{1}$ and $B_{1} C$ is $\qquad$
3. $\frac{\sqrt{6}}{6}$.

From the conditions of the cube, it is easy to know that $B_{1} C \perp$ plane $\left(A B C_{1}\right)$. Therefore, $B_{1} C$ is perpendicular to any line in $\left(A B C_{1}\right)$. Draw $Q P \perp$ $A C_{1}$ at $P$ through the intersection point $Q$ of $B_{1} C$ and $B C_{1}$. Then $B_{1} C \perp P Q$ at $Q$. $P Q$ is the distance between the skew lines $A C_{1}$ and $B_{1} C$. It is easy to find $P Q=\frac{\sqrt{6}}{6}$ from $\triangle A B C_{1} \backsim \triangle C_{1} Q P$.