$$
\begin{array}{l}
\text { ( ). } \\
m \overrightarrow{O A}+n \overrightarrow{O B}+p \overrightarrow{O C}=\mathbf{0} \text {, } \\
\text { then } S_{\triangle B O C}: S_{\triangle C O A}: S_{\triangle A O B}=m: n: p \text {; } \\
\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\mathbf{0} \text {; } \\
a \overrightarrow{O A}+b \overrightarrow{O B}+c \overrightarrow{O C}=\mathbf{0} \text {; } \\
\sin 2 A \cdot \overrightarrow{O A}+\sin 2 B \cdot \overrightarrow{O B}+\sin 2 C \cdot \overrightarrow{O C}=\mathbf{0} \text {. } \\
\end{array}
$$
6. Among the following four propositions, the number of true propositions is
(1) Point $O$ is inside $\triangle A B C$, and
(2) $O$ is the centroid of $\triangle A B C$, then
(3) $O$ is the incenter of $\triangle A B C$, then
(4) $O$ is the circumcenter of $\triangle A B C$, then
(A)1
(B) 2
(C) 3
(D) 4
$$
6. C.

First, prove: As shown in Figure 4, let $O$ be a point inside $\triangle ABC$. Then $S_{\triangle BOC} \overrightarrow{OA} + S_{\triangle COA} \overrightarrow{OB} + S_{\triangle AOB} \overrightarrow{OC} = \mathbf{0}$. (1) Note that,
$$
\begin{array}{l}
\overrightarrow{AO} = \frac{AO}{AD} \overrightarrow{AD} \\
= \frac{AO}{AD} \cdot \frac{DC}{BC} \overrightarrow{AB} + \frac{AO}{AD} \cdot \frac{BD}{BC} \overrightarrow{AC} \\
= \frac{S_{\triangle COA}}{S_{\triangle ACD}} \cdot \frac{S_{\triangle ACD}}{S_{\triangle ABC}} \overrightarrow{AB} + \frac{S_{\triangle AOB}}{S_{\triangle ABD}} \cdot \frac{S_{\triangle ABD}}{S_{\triangle ABC}} \overrightarrow{AC} \\
= \frac{S_{\triangle COA}}{S_{\triangle ABC}} \overrightarrow{AB} + \frac{S_{\triangle AOB}}{S_{\triangle ABC}} \overrightarrow{AC} \\
= \frac{S_{\triangle COA}}{S_{\triangle ABC}} (\overrightarrow{OB} - \overrightarrow{OA}) + \frac{S_{\triangle AOB}}{S_{\triangle ABC}} (\overrightarrow{OC} - \overrightarrow{OA}).
\end{array}
$$

Therefore, equation (1) holds.
It is easy to see that among the four propositions in the question, all except (1) are correct.