Given that $\angle B A D=30^{\circ} \Rightarrow B(\sqrt{3} y-1, y)$, substituting into the hyperbola equation yields $(\sqrt{3} y-1)^{2}-y^{2}=1$ $\Rightarrow y=\sqrt{3} \Rightarrow|B D|=\sqrt{3},|A D|=3 \Rightarrow S_{\triangle A B C}=3 \sqrt{3}$. Therefore, the answer is $C$.
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. $6 \sqrt{3}$