12. Given $\triangle A B C$, where $\overrightarrow{A B}=\vec{a}, \overrightarrow{A C}=\vec{b}$, express the area of $\triangle A B C$ using vector operations of $\vec{a}$ and $\vec{b}$, i.e., $S_{\triangle A B C}=$ $\qquad$
$\begin{array}{l}\text { 12. } \frac{1}{2} \sqrt{(|\vec{a}| \cdot|\vec{b}|)^{2}-(\vec{a} \cdot \vec{b})^{2}} . \\ \text { Given } S_{\triangle A B C}=\frac{1}{2}|A B| \cdot|A C| \cdot \sin A=\frac{1}{2}|\vec{a}| \cdot|\vec{b}| \cdot \sin A, \\ \text { we have } \begin{aligned} S_{\triangle A B C} & =\frac{1}{4}|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(1-\cos ^{2} A\right) \\ & =\frac{1}{4}|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left[1-\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\right)^{2}\right] \\ & =\frac{1}{4}\left[\left(|\vec{a}|^{2} \cdot|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}\right] .\right.\end{aligned}\end{array}$