Draw 63 south lines on the coordinate plane: $y=b, y=\sqrt{3} x+2 b, y=-\sqrt{3} x+2 b$, where $b=-10$, $-9,-8, \cdots, 8,9,10$. These lines divide the plane into several equilateral triangles. The number of equilateral triangles with side length $\frac{2}{\sqrt{3}}$ is $\qquad$
12. 660 .

The six outermost lines determine a regular hexagon with a side length of $\frac{20}{\sqrt{3}}$. The three lines passing through the origin $O$ divide this hexagon into six equilateral triangles with a side length of $\frac{20}{\sqrt{3}}$. Since the side length of each large triangle is 10 times that of the small triangles, and each large triangle is divided into $10^{2}$ small triangles, the number of small triangles with a side length of $\frac{2}{\sqrt{3}}$ inside the hexagon is $6 \times 10^{2}=600$ (units). Additionally, there are 10 small triangles adjacent to each side of the hexagon. Therefore, the total number of equilateral triangles with a side length of $\frac{2}{\sqrt{3}}$ is $6 \times 10^{2}+6 \times 10=660$.