3. Let $l$ and $m$ be two skew lines, and on $L$ there are three points $A$, $B$, and $C$, with $A B = B C$. Through $A$, $B$, and $C$, perpendiculars $A D$, $B E$, and $C F$ are drawn to $m$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $A D = \sqrt{15}$, $B E = \frac{7}{2}$, and $C F = \sqrt{10}$, find the distance between $l$ and $m$.
3. Let $L M$ be the line determined by lines $l$ and $m$. Suppose plane $P$ is parallel to line $l$, and through points $A, B, C$, perpendiculars $A G, B H, C K$ are drawn to plane $P$, with feet of the perpendiculars being $G, H, K$, respectively, as shown in Figure 15-12. Then points $G, H, K$ lie on a line $L^{\prime}$ parallel to $l$, and lines $C D, H E, K F$ are connected.

Since $C K \perp$ plane $P, B H \perp$ plane $P, A G \perp$ plane $P, A B = B C$, it follows that $H K = G H$.

Also, since $C F, B E, A D$ are all perpendicular to $m$, then $F K, E H, D G$ are also perpendicular to $m$. Thus, $K F \parallel E H \parallel G D$, and we have $2 E H = F K + D G$.
Let the common perpendicular distance be $x$, then $L M = C K = B H = A G = x$,
$E H = \sqrt{\frac{49}{4} - x^{2}}, F K = \sqrt{10 - x^{2}}, D G = \sqrt{15 - x^{2}}$. Therefore, we get $2 \sqrt{\frac{49}{4} - x^{2}} = \sqrt{10 - x^{2}} + \sqrt{15 - x^{2}}$, solving this yields $x = \pm \sqrt{6}$ (the negative root is discarded). Hence, the distance between $l$ and $m$ is $\sqrt{6}$.
Note: When $A, B, C$ are not all on the same side of point $L$, then $2 \sqrt{\frac{49}{4} - x^{2}} = \sqrt{15 - x^{2}} - \sqrt{10 - x^{2}}$, so $A, B, C$ must be on the same side of point $L$.