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