12. Given four points $O, A, B, C$ on a plane, satisfying $O A=4, O B=3, O C=2, \overrightarrow{O B} \cdot \overrightarrow{O C}=3$, then the maximum value of the area of $\triangle A B C$ is $\qquad$.
In $\triangle O B C$, $\angle B O C=60^{\circ} \Rightarrow B C^{2}=9+4-3 \cdot 2=7 \Rightarrow B C=\sqrt{7}$, and the distance from $O$ to $B C$ is $h=\frac{3 \sqrt{3}}{\sqrt{7}}$, so the maximum distance from $A$ to $B C$ is $h_{a}=4+\frac{3 \sqrt{3}}{\sqrt{7}}$.
Therefore, the maximum area of $\triangle A B C$ is $S=\frac{1}{2} B C \cdot h_{a}=2 \sqrt{7}+\frac{3 \sqrt{3}}{2}$.