A parallelogram is constructed on the vectors $\vec{a}= 3\vec{\alpha }-\vec{\beta }, \vec{b}=\vec{\alpha } + 3\vec{\beta} .$ If $|\vec{\alpha}|=|\vec{\beta }|=2$ and the angle between $\vec{\alpha }$ and $\vec{\beta }$ is $\frac{\pi }{3}$, then length of the diagonal of the parallelogram is :

$4\sqrt{7}$

The correct answer is Option (3) → $4\sqrt{7}$

diagonal $\vec d=\vec a+\vec b=4\vec α + \vec β$

$|\vec d|=\sqrt{\vec d.\vec d}$

$=\sqrt{16|\vec α|^2+4|\vec β|^2+16|\vec α||\vec β|\cos\frac{π}{3}}$

$=\sqrt{64+16+32}$

$=\sqrt{112}=4\sqrt{7}$