Let $l, m$ be two skew lines, on $l$ there are three points $A, B, C$, and $A B=B C$. Draw perpendiculars from $A, B, C$ to $m$, denoted 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 $L M$ be the common perpendicular of lines $l$ and $m$, with $L \in l$ and $M \in m$.
Construct a plane $P$ through $m$ parallel to line $l$, and draw perpendiculars $A G, B H, C K$ from points $A, B, C$ to plane $P$, with feet of the perpendiculars at $G, H, K$ respectively. Then points $G, H, K$ lie on a line $l^{\prime}$ parallel to $l$, and connect $G D$, $H E$, $K F$.
$$
\begin{array}{ll}
\because \quad A B=B C, A G \parallel B H \parallel C K, \\
\therefore \quad C H=H K .
\end{array}
$$

Also, since $A D \perp m, B E \perp m, C F \perp m$, by the theorem of three perpendiculars, we have
$$
G D \perp m, H E \perp m, K F \perp m .
$$

Thus, $G C \parallel H E \parallel K F$, and $E, H$ are the midpoints of $F D, K G$ respectively.
Let the distance between $l$ and $m$ be $x$, then
$$
\begin{array}{c}
A G=B H=C K=L M=x \\
\because \quad G D=\sqrt{A D^{2}-A G^{2}}, H E=\sqrt{B E^{2}-B H^{2}}, \\
K F=\sqrt{C F^{2}-C K^{2}} .
\end{array}
$$

Given $A D=\sqrt{15}, B E=\frac{7}{2}, C F=\sqrt{10}$, when points $A, B, C$ are on the same side of point $L$, we have
$$
2 H E=K F+G D
$$

Thus, $\quad 2 \sqrt{\frac{49}{4}-x^{2}}=\sqrt{10-x^{2}}+\sqrt{15-x^{2}}$
Solving this equation, we get $x= \pm \sqrt{6}$, discarding the negative root, we find that the distance between lines $l$ and $m$ is $\sqrt{6}$.