For example, the three vertices of $\triangle A B C$ are $\left(x_{1}, y_{1}\right)(i=1,2,3)$, and $P$ is any point on the plane where $\triangle A B C$ lies. If the sum of the squares of the distances from $P$ to the three vertices is minimized, find the coordinates of point $P$.

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Let $P(x, y)$, then
$$
\begin{aligned}
\sum P A^{2} & =\sum\left[\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}\right] \\
& =\sum\left(x-x_{1}\right)^{2}+\sum\left(y-y_{1}\right)^{2} \\
& =3 x^{2}-2 x \sum x_{1}+\sum x_{1}^{2}+3 y^{2}-2 y \sum y_{1}+\sum y_{1}^{2} \\
& =3\left(x-\frac{\sum x_{1}}{3}\right)^{2}+\sum x_{1}^{2}-\frac{\left(\sum x_{1}\right)^{2}}{3}+3\left(y-\frac{\sum y_{1}}{3}\right)^{2}+\sum y_{1}^{2}-\frac{\left(\sum y_{1}\right)^{2}}{3}
\end{aligned}
$$

When $x=\frac{\sum x_{1}}{3}, y=\frac{\sum y_{1}}{3}$, the above expression has its minimum value. That is, $P\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$. It is easy to see that this $P$ is the centroid of $\triangle A B C$.