As shown in Figure 1, three circles with radius $r$ are externally tangent to each other and internally tangent to $\odot O$ at points $A$, $B$, and $C$. The tangents to $\odot O$ at $A$ and $B$ intersect at point $P$. Find the area of the shaded region formed by the two tangents and the arc between the points of tangency.
Solve As shown in Figure 1, connect $O_{1} O_{2}$, $O_{2} O_{3}$, $O_{1} O_{3}$, $O A$, $O B$, $O P$, and let $O P$ intersect $\odot O$ at point $D$.

Since $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ are pairwise externally tangent, $\triangle O_{1} O_{2} O_{3}$ is an equilateral triangle with side length $2r$, and $\angle O_{2} O O_{3}=120^{\circ}$. Therefore,
$$
O O_{2}=\frac{2 \sqrt{3}}{3} r.
$$

Thus, the radius of $\odot O$, $O A=\frac{2 \sqrt{3}+3}{3} r$.
Since $\triangle P A B$ is an equilateral triangle, and
$$
\angle O P A=30^{\circ}.
$$

In the right triangle $\triangle P O A$, we have
$$
P A=(2+\sqrt{3}) r.
$$

Therefore, $S_{\triangle P O A}=\frac{1}{2} P A \cdot O A$
$$
\begin{array}{l}
=2\left[\frac{12+7 \sqrt{3}}{6} r^{2}-\frac{(7+4 \sqrt{3}) \pi r^{2}}{18}\right] \\
=\left[\frac{12+7 \sqrt{3}}{3}-\frac{(7+4 \sqrt{3}) \pi}{9}\right] r^{2}.
\end{array}
$$