Solution.

Let $O_{2} B=r$; then $O_{1} A=R=3 r$ (Fig. 10.2). Draw $O_{2} C \| A B$; we have $A C=r, O_{1} C=2 r, O_{1} O_{2}=4 r$, i.e., $O_{1} O_{2}=2 O_{1} C$ and, therefore, $\angle C O_{2} O_{1}=30^{\circ}$. Since $O_{2} C=A B=6 \sqrt{3}$, from $\triangle O_{1} C O_{2}$ we find $O_{1} O_{2}=O_{2} C: \cos 30^{\circ}=6 \sqrt{3}:(\sqrt{3} / 2)=12$, from which $r=3, R=9$. The arcs entering the specified figure contain 120 and $240^{\circ}$, respectively, so their lengths are $\frac{2 \pi r}{3}$ and $\frac{4 \pi r}{3}$. The desired perimeter is $P=2 \cdot 6 \sqrt{3}+\frac{2 \pi r}{3}+\frac{4 \pi r}{3}=12 \sqrt{3}+14 \pi$.

Answer: $14 \pi+12 \sqrt{3}$.
10.192. Two circles touch each other externally. Their radii are in the ratio 3:1, and the length of their common external tangent is $6 \sqrt{3}$. Determine the perimeter of the figure formed by the external tangents and the external parts of the circles.