Example: Given the radii of the upper and lower bases of a frustum are 3 and 6, respectively, and the height is $3 \sqrt{3}$, the radii $O A$ and $O B$ of the lower base are perpendicular, and $C$ is a point on the generatrix $B B^{\prime}$ such that $B^{\prime} C: C B$ $=1: 2$. Find the shortest distance between points $A$ and $C$ on the lateral surface of the frustum.
Given $B^{\prime} D \perp O B$, then $B^{\prime} D=3 \sqrt{3}$.
$$
\begin{array}{l}
\because O^{\prime} B^{\prime}=3, O B=6, \\
\therefore B^{\prime} B=\sqrt{(3 \sqrt{3})^{2}+(6-3)^{2}}=6,
\end{array}
$$

and $O^{\prime} B^{\prime}: O B=3: 6=1: 2$,
$\therefore$ In the development diagram, $B^{\prime}$ is the midpoint of $V B$, $V B=12$. Given $B^{\prime} C: C B=1: 2$,
$$
\begin{array}{c}
\therefore B^{\prime} C=\frac{1}{3} B^{\prime} \cdot B=2, \\
V C=6+2=8 .
\end{array}
$$

Let the central angle of the sector ring in the frustum's lateral development be $\theta$.
$$
\begin{array}{l}
\theta=\frac{6-3}{6} \times 360^{\circ}=180^{\circ}, \\
\alpha=\frac{1}{4} \theta=45^{\circ} .
\end{array}
$$

In $\triangle V A C$,
$$
\begin{aligned}
A C^{2} & =V A^{2}+V C^{2}-2 V A \cdot V C \cos \alpha \\
& =12^{2}+8^{2}-2 \times 12 \times 8 \times \cos 45^{\circ} \\
& =208-96 \sqrt{2}, \\
\therefore A C & =4 \sqrt{13-6 \sqrt{2}} .
\end{aligned}
$$