(12) Let the acute triangle $ABC$ have internal angles $A, B, C$ opposite to sides $a, b, c$ respectively, and $\sin A = 2a \sin B$, with the circumradius of $\triangle ABC$ being $\frac{\sqrt{3}}{6}$. Then the range of the perimeter of $\triangle ABC$ is . $\qquad$
12 Because $\sin A=2 a \sin B$, by the Law of Sines we know
$$
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R,
$$

where $R$ is the circumradius of $\triangle ABC$. Therefore, $\frac{a}{2 R}=2 a \cdot \frac{b}{2 R}$, which implies $b=\frac{1}{2}$. And
$$
\sin B=\frac{b}{2 R}=\frac{\frac{1}{2}}{2 \cdot \frac{\sqrt{3}}{6}}=\frac{\sqrt{3}}{2},
$$

Since $B \in\left(0, \frac{\pi}{2}\right)$, we have $B=\frac{\pi}{3}$. Let the perimeter of $\triangle ABC$ be $l$. Because
$$
A+C=\pi-\frac{\pi}{3}=\frac{2 \pi}{3},
$$

we have $C=\frac{2 \pi}{3}-A$. Since $\triangle ABC$ is an acute triangle, $A \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$. Therefore,
$$
\begin{aligned}
l & =a+b+c \\
& =2 R \sin A+2 R \sin C+b \\
& =\frac{1}{2}+\frac{\sqrt{3}}{3}(\sin A+\sin C) \\
& =\frac{1}{2}+\frac{\sqrt{3}}{3}\left(\sin A+\sin \left(\frac{2}{3} \pi-A\right)\right) \\
& =\frac{1}{2}+\frac{\sqrt{3}}{3}\left(\sin A+\sin \frac{2}{3} \pi \cos A-\cos \frac{2}{3} \pi \sin A\right) \\
& =\frac{1}{2}+\frac{\sqrt{3}}{3}\left(\frac{\sqrt{3}}{2} \cos A+\frac{3}{2} \sin A\right) \\
& =\frac{1}{2}+\left(\frac{1}{2} \cos A+\frac{\sqrt{3}}{2} \sin A\right) \\
& =\frac{1}{2}+\sin \left(A+\frac{\pi}{6}\right),
\end{aligned}
$$

Since $A \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$, then $A+\frac{\pi}{6} \in\left(\frac{\pi}{3}, \frac{2 \pi}{3}\right)$, thus
$$
\begin{array}{c}
\sin \left(A+\frac{\pi}{6}\right) \in\left(\frac{\sqrt{3}}{2}, 1\right], \\
\frac{1}{2}+\sin \left(A+\frac{\pi}{6}\right) \in\left(\frac{1+\sqrt{3}}{2}, \frac{3}{2}\right],
\end{array}
$$

Therefore, the range of the perimeter of $\triangle ABC$ is $\left(\frac{1+\sqrt{3}}{2}, \frac{3}{2}\right]$, and the maximum perimeter is $\frac{3}{2}$ when $A=\frac{\pi}{3}$, i.e., when $\triangle ABC$ is an equilateral triangle.