11. On a sphere with radius $R$, there are four points $A, B, C, D$, such that $A B=B C=C A=3$. If the maximum volume of the tetrahedron $A B C D$ is $\frac{9 \sqrt{3}}{4}$, then the surface area of this sphere is $\qquad$ .
As shown in the figure, $O_{1} C=\sqrt{3} \Rightarrow O O_{1}=\sqrt{R^{2}-3}$. Since $S_{\triangle A B C}=\frac{9 \sqrt{3}}{4}$, then $h=R+\sqrt{R^{2}-3} \leqslant 3$ $\Rightarrow R \leqslant 2$. Therefore, when the maximum volume of the tetrahedron $A B C D$ is $\frac{9 \sqrt{3}}{4}$, the surface area of this sphere is $16 \pi$.