17. As shown in the figure, it is known that the three vertices of $\triangle A B C$ are on the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, and the coordinate origin $O$ is the centroid of $\triangle A B C$. Try to find the area of $\triangle A B C$.
Given $A(2 \sqrt{3} \cos \theta, 2 \sin \theta)$, connect $A O$ to intersect $B C$ at point $D$,
then $D(-\sqrt{3} \cos \theta, -\sin \theta)$, and by the conclusion of the point difference method,
$$
k_{B C} \cdot \frac{y_{D}}{x_{D}}=-\frac{1}{3} \Rightarrow k_{B C}=-\frac{\sqrt{3}}{3 \tan \theta} \text {. }
$$

Thus, the line $B C: y+\sin \theta=-\frac{\sqrt{3}}{3 \tan \theta}(x+\sqrt{3} \cos \theta)$
$$
\Rightarrow x \cos \theta+\sqrt{3} y \sin \theta+\sqrt{3}=0 \text {. }
$$

Solving the system of equations $\left\{\begin{array}{l}x \cos \theta+\sqrt{3} y \sin \theta+\sqrt{3}=0, \\ x^{2}+3 y^{2}=12\end{array} \Rightarrow x^{2}+2 \sqrt{3} x \cos \theta+3-12 \sin ^{2} \theta=0\right.$.
Thus, $|B C|=\sqrt{1+\frac{1}{3 \tan ^{2} \theta}} \cdot \sqrt{12 \cos ^{2} \theta-4\left(3-12 \sin ^{2} \theta\right)}$
$=\frac{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}}{|\sqrt{3} \sin \theta|} \cdot|6 \sin \theta|=2 \sqrt{3} \cdot \sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}$, and the distance from point $A$ to line $B C$ is
$$
d=\frac{\left|2 \sqrt{3} \cos ^{2} \theta+2 \sqrt{3} \sin ^{2} \theta+\sqrt{3}\right|}{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}}=\frac{3 \sqrt{3}}{\sqrt{\cos ^{2} \theta+3 \sin ^{2} \theta}},
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2}|B C| \cdot d=\frac{1}{2} \cdot 2 \sqrt{3} \cdot 3 \sqrt{3}=9$.