9. Given four points $O, A, B, C$ on a plane, satisfying
$$
O A=4, O B=3, O C=2, \overrightarrow{O B} \cdot \overrightarrow{O C}=3 \text {. }
$$

Then the maximum area of $\triangle A B C$ is $\qquad$
$9.2 \sqrt{7}+\frac{3 \sqrt{3}}{2}$.
From the problem, we know
$$
B C=\sqrt{O B^{2}+O C^{2}-2 \overrightarrow{O B} \cdot \overrightarrow{O C}}=\sqrt{7} \text {, }
$$

and the angle between $O B$ and $O C$ is $60^{\circ}$.
Consider three circles $\Gamma_{1} 、 \Gamma_{2} 、 \Gamma_{3}$ centered at the origin $O$ with radii $2,3,4$ respectively. Then, points $C 、 B$ can be fixed on circles $\Gamma_{1}$ and $\Gamma_{2}$, and point $A$ can move on circle $\Gamma_{3}$. Draw $O D \perp B C$ at point $D$. The area of $\triangle A B C$ reaches its maximum value if and only if points $D 、 O 、 A$ are collinear and $\overrightarrow{D O}$ is in the same direction as $\overrightarrow{D A}$. At this time, we have
$$
\begin{array}{l}
B C \cdot O D=O C \cdot O B \sin \angle B O C \\
\Rightarrow O D=\frac{3 \sqrt{21}}{7} . \\
\text { Therefore, }\left(S_{\triangle A B C}\right)_{\max }=\frac{(O A+O D) B C}{2} \\
=\frac{1}{2}\left(4+\frac{3 \sqrt{21}}{7}\right) \times \sqrt{7}=2 \sqrt{7}+\frac{3 \sqrt{3}}{2} .
\end{array}
$$