[Example 1.3.8] Let $l, m$ be two skew lines, and on $l$ there are three points $A, B, C$ such that $AB = BC$. Through $A, B, C$ respectively, draw perpendiculars to $m$ at points $D, E, F$. It is known that $AD = \sqrt{15}, BE = \frac{7}{2}, CF = \sqrt{10}$. Find the distance between $l$ and $m$.
Let the common perpendicular of skew lines $l$ and $m$ be $LM$, and let $LM = x$.
Construct a plane $\alpha$ through line $m$ parallel to line $l$.
We aim to transform the problem of finding $x$ into a calculation problem within the plane $\alpha$.

For this purpose, draw perpendiculars $AG, BH, CK$ from points $A, B, C$ to plane $\alpha$, with feet at $G, H, K$ respectively.
Then $G, H, K$ lie on line $l'$, and $l' \parallel l, l' \subset \alpha$.
Connect $GD, HE, KF$. By
$$
AB = BC, AG \parallel BH \parallel CK
$$

we get
$$
GH = HK,
$$

and by the inverse of the Three Perpendiculars Theorem,
$$
GD \perp m, HE \perp m, KF \perp m.
$$

Furthermore,

Thus,
$$
\begin{array}{c}
AG = BH = CK = LM = x. \\
GD = \sqrt{AD^2 - AG^2} = \sqrt{15 - x^2}, \\
HE = \sqrt{BE^2 - BH^2} = \sqrt{\frac{49}{4} - x^2}, \\
KF = \sqrt{CF^2 - CK^2} = \sqrt{10 - x^2}.
\end{array}
$$

Since $KFDG$ is a trapezoid and $HE$ is the midline of the trapezoid, we have
$$
\begin{array}{c}
HE = \frac{GD + KF}{2}, \\
\sqrt{\frac{49}{4} - x^2} = \frac{\sqrt{15 - x^2} + \sqrt{10 - x^2}}{2},
\end{array}
$$

Solving this irrational equation, and given $x > 0$, we get
$$
x = \sqrt{6}.
$$

There is another case where points $A, B, C$ are not on the same side of $L$, for example, $C$ is on the opposite side of $A, B$ with respect to $L$. In this case,
$$
HE = \frac{GD - KF}{2},
$$

the corresponding irrational equation is
$$
\sqrt{\frac{49}{4} - x^2} = \frac{\sqrt{15 - x^2} + \sqrt{10 - x^2}}{2},
$$

which has no solution.
Therefore, the distance between the skew lines $l$ and $m$ is $\sqrt{6}$.
This problem is solved by projecting elements on $l$ to elements on $l'$ in $\alpha$, transforming the spatial calculation problem into a planar problem using the Three Perpendiculars Theorem and the Pythagorean Theorem, and then solving the equation to complete the solution.