一、(20 points) Given that the radii of $\odot A$, $\odot B$, and $\odot C$ are all 3, and they are mutually externally tangent. On the plane where these three circles lie, take all points that are a distance of $2 \sqrt{3} - 3$ from at least one point on these circles. Find the area of the region formed by all such points.
Given that $\triangle ABC$ is an equilateral triangle with a side length of 6.
Let the center of $\triangle ABC$ be $O$. Then
$$
O A=O B=O C=2 \sqrt{3} \text{. }
$$

As shown in Figure 2, construct the reflections of point $O$ about the sides $AB$, $BC$, and $CA$ as points $O_{1}$, $O_{2}$, and $O_{3}$, respectively. Connect $A O_{1}$, $O_{1} B$, $B O_{2}$, $O_{2} C$, $C O_{3}$, and $O_{3} A$.

Thus, the desired region consists of three sectors $O_{1} B O_{2}$, $O_{2} C O_{3}$, and $O_{3} A O_{1}$ (with radius $2 \sqrt{3}$ and central angle $\frac{4 \pi}{3}$) and the regular hexagon $A O_{1} B O_{2} C O_{3}$ minus the areas of three circles $\odot A$, $\odot B$, and $\odot C$ (with radius $6-2 \sqrt{3}$). The area is
$$
\begin{array}{l}
3 \times \frac{2 \pi}{3}(2 \sqrt{3})^{2}+6 \times \frac{\sqrt{3}}{4}(2 \sqrt{3})^{2}-3 \pi(6-2 \sqrt{3})^{2} \\
=24 \pi+18 \sqrt{3}-36 \pi(4-2 \sqrt{3}) \\
=18 \sqrt{3}+\pi(72 \sqrt{3}-120) .
\end{array}
$$