Example 3 Let the coordinates of the three vertices of the equilateral $\triangle A B C$ be $\left(x_{A}, y_{A}\right),\left(x_{B}, y_{B}\right),\left(x_{C}, y_{C}\right)$. Please express $S_{\triangle A B C}$ using $x_{A}, x_{B}, x_{C}$.
【Analysis】Let $\triangle A B C$ be counterclockwise, and $O$ be the center of the equilateral $\triangle A B C$, with $R$ being the radius of the circumcircle. Then
$$
x_{0}=\frac{x_{A}+x_{B}+x_{C}}{3} .
$$

Therefore, the x-coordinate of vector $\overrightarrow{O A}$ is
$$
x_{A}-x_{o}=\frac{2 x_{A}-x_{B}-x_{C}}{3}=R \cos \varphi \text {, }
$$

the x-coordinate of vector $\overrightarrow{O B}$ is
$$
x_{B}-x_{o}=R \cos \left(\varphi+120^{\circ}\right) \text {, }
$$

and the x-coordinate of vector $\overrightarrow{O C}$ is
$$
\begin{array}{l}
x_{C}-x_{o}=R \cos \left(\varphi-120^{\circ}\right) . \\
\text { Hence }\left(x_{C}-x_{O}\right)-\left(x_{B}-x_{O}\right) \\
=R\left[\cos \left(\varphi-120^{\circ}\right)-\cos \left(\varphi+120^{\circ}\right)\right] \\
\Rightarrow x_{C}-x_{B}=2 R \sin \varphi \cdot \sin 120^{\circ}=\sqrt{3} R \sin \varphi . \\
\text { Then }\left[\frac{\left(2 x_{A}-x_{B}-x_{C}\right)}{3}\right]^{2}+\left(\frac{x_{C}-x_{B}}{\sqrt{3}}\right)^{2} \\
=R^{2} \cos ^{2} \varphi+R^{2} \sin ^{2} \varphi=R^{2} \\
\Rightarrow R^{2}=\frac{4}{9} \sum x_{A}^{2}-\frac{4}{9} \sum x_{A} x_{B},
\end{array}
$$

where “$\sum$” denotes the cyclic sum.
Then $S_{\triangle A B C}=3 S_{\triangle A O B}=3 \times \frac{1}{2} R^{2} \sin 120^{\circ}$
$$
=\frac{3 \sqrt{3}}{4} R^{2}=\frac{\sqrt{3}}{3} \sum\left(x_{A}^{2}-x_{B} x_{C}\right) .
$$