3. Given that $A, B, C$ are three non-collinear points on a plane, $O$ is the centroid of $\triangle A B C$, and a moving point $P$ satisfies
$$
O P=\frac{1}{3}\left(\frac{1}{2} O A+\frac{1}{2} O B+2 O C\right) .
$$

Then $P$ must be the ( ) of $\triangle A B C$.
(A) midpoint of the median from side $A B$
(B) one of the trisection points (not the centroid) of the median from side $A B$
(C) centroid
(D) midpoint of side $A B$
3. B.

As shown in Figure 2, since $O$ is the centroid of $\triangle A B C$, we have:
$$
\begin{aligned}
& O A+O B+O C \\
= & 0 . \\
& \text { Therefore, } O P \\
= & \frac{1}{3}\left(\frac{1}{2} O A+\frac{1}{2} O B+2 O C\right) \\
= & \frac{1}{6}(O A+O B+4 O C) \\
= & \frac{1}{6}(-O C+4 O C)=\frac{1}{2} O C .
\end{aligned}
$$

Thus, $P$ is at the point that trisects the median $C F$ of $A B$ (not the centroid).