12.1.7 Let $l$ and $m$ be two skew lines, and on $l$ there are three points $A$, $B$, and $C$, such that $AB = BC$. Through $A$, $B$, and $C$, perpendiculars $AD$, $BE$, and $CF$ are drawn to $m$, with feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $AD = \sqrt{15}$, $BE = \frac{7}{2}$, and $CF = \sqrt{10}$, find: the distance between $l$ and $m$.
As shown in the figure, let $LM$ be the common perpendicular of lines $l$ and $m$, with $L \in l$ and $M \in m$. Construct a plane $P$ parallel to line $l$ through $m$, and draw perpendiculars $AG$, $BH$, and $CK$ from points $A$, $B$, and $C$ to plane $P$, with feet of the perpendiculars at $G$, $H$, and $K$, respectively. Then points $G$, $H$, and $K$ lie on a line $l'$ parallel to $l$. Connect $GD$, $HE$, and $KF$. Since $AB = BC$ and $AG \parallel BH \parallel CK$, it follows that $GH = HK$. Also, since $AD \perp m$, $BE \perp m$, and $CF \perp m$, by the inverse of the three perpendiculars theorem, we have $GD \perp m$, $HE \perp m$, and $KF \perp m$. Therefore, $GD \parallel HE \parallel KF$, and points $E$ and $H$ are the midpoints of $FD$ and $KG$, respectively.

Let the distance between $l$ and $m$ be $x$, then $AG = BH = CK = LM = x$. When points $A$, $B$, and $C$ are on the same side of point $L$, we have $2HE = KF + GD$. By $GD = \sqrt{AD^2 - AG^2}$, $HE = \sqrt{BE^2 - BH^2}$, and $KF = \sqrt{CF^2 - CK^2}$, and given $AD = \sqrt{15}$, $BE = \frac{7}{2}$, and $CF = \sqrt{10}$, we get $2 \sqrt{\frac{49}{4} - x^2} - \sqrt{10 - x^2} + \sqrt{15 - x^2}$. Solving this equation, we find $x = \sqrt{6} (x > 0)$. Therefore, the distance between lines $l$ and $m$ is $\sqrt{6}$.