3. (2000 National High School Competition Question) Given that $A$ 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
3. C. Let $B$ be above the $x$-axis, then $k_{A B}=\tan 30^{\circ}=\frac{\sqrt{3}}{3},$ the equation of $A B$ is $y=\frac{\sqrt{3}}{3}(x+1)$. Substituting into the hyperbola equation $x^{2}-$ $y^{2}=1$, we get the coordinates of point $B$ as $(2, \sqrt{3})$. Similarly, we can get $C(2,-\sqrt{3})$, so $S_{\triangle A B C}=[(2-(-1)] \sqrt{3}=3 \sqrt{3}$, hence the answer is $C$.