12. Let the side length of the equilateral $\triangle ABC$ be $a$. If $\triangle ABC$ is rotated $60^{\circ}$ about its center (the center of the circumcircle of the equilateral triangle), the resulting triangle is $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then the distance between points $A$ and $B^{\prime}$ is $\qquad$ .
12. $\frac{\sqrt{3}}{3} a$ or $\frac{2 \sqrt{3}}{3} a$.

When $\triangle A B C$ is rotated $60^{\circ}$ clockwise, as shown in Figure 13
(a), connect $O A$ and $O B^{\prime}$. Since $\angle A O B^{\prime}=60^{\circ}$, $\triangle O A B^{\prime}$ is an equilateral triangle. Therefore, $A B^{\prime}=\frac{\sqrt{3}}{3} a$.
When $\triangle A B C$ is rotated $60^{\circ}$ counterclockwise, as shown in Figure 13
(b), $A B^{\prime}=2 O A=\frac{2 \sqrt{3}}{3} a$.