14. 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 $m$.
Solution
As shown in the figure, let the common perpendicular segment of the skew lines $l, m$ be
$MN$, draw $l_{1} / / l$ through $N$, and draw $A A_{1}, B B_{1}$,
$C C_{1}$ parallel to $MN$, intersecting $l_{1}$ at $A_{1}, B_{1}, C_{1}$,
connect $A_{1} D, B_{1} E, C_{1} F$.
Let $A_{1} D=y_{1}, C_{1} F=y_{2}, B_{1} E=y$,
$C_{1} M=m, M B_{1}=n, B_{1} A_{1}=m+n$,
from $\triangle C_{1} M F \sim \triangle B_{1} M E \Rightarrow \frac{m}{n}=\frac{y_{2}}{y}$,
$\triangle B_{1} M E \sim \triangle A_{1} M D \Rightarrow \frac{m+2 n}{n}=\frac{y_{1}}{y} \Rightarrow \frac{m}{n}=\frac{y_{1}-2 y}{y}$,
from (1), (2) we get $y_{2}=y_{1}-2 y \Rightarrow 2 y=y_{1}-y_{2}$. Let $A A_{1}=B B_{1}=C C_{1}=M N=x$,
thus $2 \sqrt{\frac{49}{4}-x^{2}}=\sqrt{15-x^{2}}-\sqrt{10-x^{2}}$
$\Rightarrow 49-4 x^{2}=25-2 x^{2}-2 \sqrt{\left(15-x^{2}\right)\left(10-x^{2}\right)} \Rightarrow \sqrt{\left(15-x^{2}\right)\left(10-x^{2}\right)}=x^{2}-12$
$\Rightarrow x^{4}-25 x^{2}+150=x^{4}-24 x^{2}+144 \Rightarrow x=\sqrt{6}$. Therefore, the distance between $l$ and $m$ is $\sqrt{6}$.