10. Let $l, m$ be two skew lines, on $l$ there are three points $A, B, C$, and $A B=B C$. Through $A, B, C$ respectively, draw perpendiculars to $m$ as $A D, B E, C F$, with feet of the perpendiculars being $D, E, F$ respectively. Given $A D=\sqrt{15}, B E=\frac{7}{2}, C F=\sqrt{10}$, find the distance between $l$ and $m$.
Let the distance between the skew lines $l$ and $m$ be $x$, then $A G = B H = C K = L M = x$.
(1) When $A, B, C$ are on the same side of point $L$, $2 H E = K F + G D$,

Since
$$
\begin{array}{l}
H E = \sqrt{B E^{2} - B H^{2}} = \sqrt{\frac{49}{4} - x^{2}}, \\
K F = \sqrt{C F^{2} - C K^{2}} = \sqrt{10 - x^{2}}, \\
G D = \sqrt{A D^{2} - A G^{2}} = \sqrt{15 - x^{2}} .
\end{array}
$$

Thus, $2 \sqrt{\frac{49}{4} - x^{2}} = \sqrt{15 - x^{2}} + \sqrt{10 - x^{2}}$,
Solving for $x$ gives $x = \pm \sqrt{6}$. Discarding the negative root, we get $x = \sqrt{6}$, which means the distance between $l$ and $m$ is $\sqrt{6}$.
(2) When $A, B, C$ are not all on the same side of point $L$, $2 H E = |K F - G D|$,

That is, $2 \sqrt{\frac{49}{4} - x^{2}} = \left| \sqrt{15 - x^{2}} - \sqrt{10 - x^{2}} \right|$
In this case, the equation has no solution, so $A, B, C$ must be on the same side of point $L$.
Therefore, the distance between $l$ and $m$ is $\sqrt{6}$.