3. Given point $A$ is the left vertex of the hyperbola $x^{2}-y^{2}=1$, points $B$ and $C$ are on the right branch of the hyperbola, and $\triangle A B C$ 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. (C).

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

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 $C$ are $(2,-\sqrt{3})$. Therefore, the area of $\triangle ABC$ is $[2-(-1)] \sqrt{3}=3 \sqrt{3}$.