3. Given point $A$ is the left vertex of the hyperbola $x^{2}-y^{2}=1$, and points $B$ and $C$ are on the right branch of the hyperbola, $\triangle A B C$ is an equilateral triangle. Then the area of $\triangle A B C$ is ( ).
A. $\frac{\sqrt{3}}{3}$
B. $\frac{3 \sqrt{3}}{2}$
C. $3 \sqrt{3}$
D. $6 \sqrt{3}$
3. $\mathrm{C}$

Let point $B$ be above the $x$-axis. Since $\triangle A B C$ is an equilateral triangle, the slope of line $A B$ is $k=\frac{\sqrt{3}}{3}$. The line also passes through point $A(-1,0)$, so the equation is
$$
y=\frac{\sqrt{3}}{3} x+\frac{\sqrt{3}}{3} .
$$

Substituting into the hyperbola equation $x^{2}-y^{2}=1$, we get the coordinates of point $B$ as $(2, \sqrt{3})$.
Similarly, the coordinates of point $\mathrm{C}$ are $(2,-\sqrt{3})$.
Thus, the area of $\triangle A B C$ is $[2-(-1)] \sqrt{3}=3 \sqrt{3}$.