(3) Given that $A, B, C$ are three non-collinear points on a plane, $O$ is the centroid of triangle $ABC$, and a moving point $P$ satisfies $\overrightarrow{O P}=\frac{1}{3} \cdot\left(\frac{1}{2} \overrightarrow{O A}+\frac{1}{2} \overrightarrow{O B}+2 \overrightarrow{O C}\right)$, then point $P$ must be the ( ) of triangle $ABC$.
(A) midpoint of the median from $A B$
(B) trisection point of the median from $A B$ (not the centroid)
(C) centroid
(D) midpoint of $A B$
(3) Since $O$ is the centroid of $\triangle A B C$, we have $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}$. Therefore,
$$
\begin{aligned}
\overrightarrow{O P} & =\frac{1}{3} \cdot\left(\frac{1}{2} \overrightarrow{O A}+\frac{1}{2} \overrightarrow{O B}+2 \overrightarrow{O C}\right) \\
& =\frac{1}{6} \cdot(\overrightarrow{O A}+\overrightarrow{O B}+4 \overrightarrow{O C}) \\
& =\frac{1}{6} \cdot(-\overrightarrow{O C}+4 \overrightarrow{O C}) \\
& =\frac{1}{2} \overrightarrow{O C} .
\end{aligned}
$$

As shown in the figure, $P$ is at the point that trisects the median $C F$ of $A B$ (not the centroid). Therefore, the answer is B.