12. Given four points $O, A, B, C$ on a plane, satisfying $OA=4, OB=3, OC=2, \overrightarrow{OB} \cdot \overrightarrow{OC}=3$. Then the maximum value of $S_{\triangle ABC}$ is $\qquad$
12. $2 \sqrt{7}+\frac{3 \sqrt{3}}{2}$.

From the given information,
$$
B C=\sqrt{O B^{2}+O C^{2}-2 \overrightarrow{O B} \cdot \overrightarrow{O C}}=\sqrt{7} \text {, }
$$

and the angle between $O B$ and $O C$ is $60^{\circ}$.
Consider a circle centered at the origin $O$ with radii 2, 3, and 4, denoted as $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$, respectively. Then, we can fix points $C$ and $B$ on $\odot O_{1}$ and $\odot O_{2}$, and let point $A$ move on $\odot O_{3}$. 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 $D$, $O$, and $A$ are collinear, and $\overrightarrow{D O}$ and $\overrightarrow{D A}$ have the same direction. At this time,
$$
\begin{array}{l}
\text { by } B C \cdot O D=O C \cdot O B \sin \angle B O C=2 \sqrt{3} \\
\Rightarrow O D=\frac{3 \sqrt{21}}{7} \\
\Rightarrow\left(S_{\triangle A B C}\right)_{\max }=\frac{(O A+O D) B C}{2} \\
\quad=2 \sqrt{7}+\frac{3 \sqrt{3}}{2} .
\end{array}
$$