11. The answer is $\mathbf{( E )}$. Let $O, O_{1}, O_{2}$ be the centers of $\Gamma, \Gamma_{1}, \Gamma_{2}$ respectively. Since $\Gamma_{1}, \Gamma_{2}$ are externally tangent, the distance $O_{1} O_{2}$ is equal to the sum of the radii of $\Gamma_{1}, \Gamma_{2}$, i.e., $O_{1} O_{2}=5$. Similarly, the distance $O O_{1}$ is equal to the difference of the radii of $\Gamma$ and $\Gamma_{1}$ (and is thus equal to $6-3=3$), and the distance $O O_{2}$ is equal to $6-2=4$. Since $O O_{2}^{2}+O O_{1}^{2}=3^{2}+4^{2}=5^{2}=O_{1} O_{2}^{2}$, the triangle $O O_{1} O_{2}$ is a right triangle at $O$, and the triangle $\mathrm{A}_{1} \mathrm{OA}_{2}$ is a right isosceles triangle.

Let $O^{\prime}$ be the center of the circumcircle of $A_{1} A A_{2}$. The triangle $A_{1} O^{\prime} A_{2}$ is certainly isosceles on base $A_{1} A_{2}$; we want to show that it is also a right triangle at $O^{\prime}$, i.e., that $O^{\prime}$ is the reflection of $O$ with respect to $A_{1} A_{2}$, hence that $O^{\prime} A_{1}=O^{\prime} A_{2}=6$.

We have $\widehat{A_{1} O^{\prime} A_{2}}=2\left(180^{\circ}-\widehat{A_{1} A A_{2}}\right)$ (central angle that subtends the same arc as $\widehat{A_{1} A A_{2}}$, but on the opposite side); on the other hand, a quick calculation of angles gives $180^{\circ}-\widehat{A_{1} A A_{2}}=$ $\widehat{A_{1} A O_{1}}+\widehat{A_{2} A O_{2}}=\widehat{A A_{1} O_{1}}+\widehat{A A_{2} O_{2}}=\widehat{A O_{1} O} / 2+\widehat{A O_{2} O} / 2=45^{\circ}$, which precisely implies the desired orthogonality.
11. Three circles $\Gamma, \Gamma_{1}, \Gamma_{2}$ have radii 6, 3, 2, respectively. $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at $A$, while $\Gamma$ is internally tangent to both other circles, respectively at $A_{1}$ and $A_{2}$. Determine the radius of the circumcircle of $A A_{1} A_{2}$.
(A) $2 \sqrt{6}$
(B) 5
(C) $\sqrt{2}+\sqrt{3}+\sqrt{6}$
(D) $4+\sqrt{3}$
(E) 6