7. In tetrahedron $ABCD$, $AB=1, BC=5, CD$ $=7, DA=5, AC=5, BD=2\sqrt{6}$. Then the distance between the skew lines $AC$ and $BD$ is $\qquad$
7. $\frac{3 \sqrt{11}}{10}$.

Given $A B^{2}+C D^{2}=50=A D^{2}+B C^{2}$
$$
\Rightarrow A C \perp B D \text {. }
$$

As shown in Figure 3, draw $B E \perp A C$ at point $E$. Then $D E \perp A C$.
Let $A E=x, B E=y$.
By $x^{2}+E B^{2}=1^{2}$,
$$
y^{2}+E B^{2}=5^{2} \text {, }
$$

we have
$$
\begin{aligned}
& y^{2}-x^{2}=24 . \\
& \text { Also } x+y=5 \text {, then } \\
& y-x=\frac{24}{5} \\
\Rightarrow & 2 x=\frac{1}{5} \\
\Rightarrow & x=\frac{1}{10} \\
\Rightarrow & B E^{2}=\frac{99}{100}, D E^{2}=\frac{2499}{100} \\
\Rightarrow & D E^{2}-B E^{2}=24=B D^{2} \\
\Rightarrow & B E \perp B D .
\end{aligned}
$$

Therefore, $B E$ is the common perpendicular segment of the skew lines $A C$ and $B D$. Thus, $B E=\frac{3 \sqrt{11}}{10}$.