Example 8 A point $P$ in space is at distances 2 and 3 from the vertices $A, B$ of a regular tetrahedron $ABCD$. When the edge length and position of the tetrahedron change, find the maximum distance from point $P$ to the line $CD$, and the length of $AB$ at this time.
Solution: (1) Once the tetrahedron is determined, the trajectory of point $P$ is also determined. The plane of the trajectory is perpendicular to $AB$. Take the midpoint $E$ of $CD$. Only when point $P$ is located in the plane $ABE$ and on the same side of $AB$ as point $E$, can the distance from $P$ to $CD$ reach its maximum value. Thus, the problem is transformed into a planar problem: "Given that the distances from point $P$ to the vertices $A$ and $B$ of $\triangle ABE$ are $2$ and $3$ respectively, and $AE = BE = \frac{\sqrt{3}}{2} AB$. When the side lengths and position of $\triangle ABE$ change, find the maximum distance from point $P$ to point $E$."
(2) As shown in Figure $16-10$, construct $\angle P^{\prime}AE = \angle PAB$ such that $P^{\prime}A = \frac{\sqrt{3}}{2} AP$.

Since $\frac{P^{\prime}A}{PA} = \frac{AE}{AB} = \frac{\sqrt{3}}{2}$ and $\angle PAB = \angle P^{\prime}AE$, we have $\triangle APB \sim \triangle AP^{\prime}E$, thus
$$
P^{\prime}E = \frac{\sqrt{3}}{2} PB = \frac{3\sqrt{3}}{2}.
$$

Also, $\frac{AP}{AB} = \frac{AP^{\prime}}{AE}$ and $\angle PAP^{\prime} = \angle BAE$, so $\triangle APP^{\prime} \sim \triangle ABE$, thus $PP^{\prime} = \frac{\sqrt{3}}{2} PA = \sqrt{3}$. Therefore, $PE \leq PP^{\prime} + P^{\prime}E = \frac{5\sqrt{3}}{2}$.
(3) When $PE$ reaches its maximum value, $P^{\prime}, P, E$ are collinear. Given $PA = 2$, $\angle APE = \angle ABE$, and $PE = \frac{5\sqrt{3}}{2}$, we can get $\cos \angle APE = \frac{\sqrt{3}}{3}$, $AE = \frac{\sqrt{51}}{2}$, and $AB = \frac{\sqrt{153}}{3}$.