For a triangle A B C, let $\vec{p}=\overrightarrow{B C}, \vec{q}=\overrightarrow{C A}$ and $\vec{r}=\overrightarrow{B A}$. If $|\vec{p}|=2 \sqrt{3},|\vec{q}|=2$ and $\cos \theta=\frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times(\vec{q}-3 \vec{r})|^2+3|\vec{r}|^2$ is equal to :