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Question Id : 17924 | Context : JEE Main 2026 (21 January Evening Shift)
For a triangle $ABC$, let $\vec{p} = \overrightarrow{BC},; \vec{q} = \overrightarrow{CA}$ and $\vec{r} = \overrightarrow{BA}$. 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:


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