3. The point that minimizes the sum of the squares of the distances to the three vertices of a triangle is the ( ) of the triangle.
(A) orthocenter
(B) centroid
(C) circumcenter
(D) incenter
3. B.

Let the triangle be $\triangle A B C$, with its centroid as $G$, and $P$ as any point on the plane. Then,
$$
\begin{aligned}
& P A^{2}+P B^{2}+P C^{2} \\
= & (\boldsymbol{P G}+\boldsymbol{G A})^{2}+(\boldsymbol{P G}+\boldsymbol{G B})^{2}+(\boldsymbol{P G}+\boldsymbol{G C})^{2} \\
= & \boldsymbol{G A}+\boldsymbol{G} \boldsymbol{B}^{2}+\boldsymbol{G} \boldsymbol{C}^{2}+3 P \boldsymbol{G}^{2}+ \\
& 2 \boldsymbol{P G}(\boldsymbol{G A}+\boldsymbol{G B}+\boldsymbol{G C}) .
\end{aligned}
$$

By the property of the centroid, $\boldsymbol{G A}+\boldsymbol{G B}+\boldsymbol{G C}=0$, thus
$$
\begin{array}{l}
P A^{2}+P B^{2}+P C^{2}=G A^{2}+G B^{2}+G C^{2}+3 P G^{2} \\
\geqslant G A^{2}+G B^{2}+G C^{2} .
\end{array}
$$

Equality holds if and only if point $P$ is the centroid $G$.