If $|a|=2$ and $|b|=3$ and the angle between $a$ and $b$ is $120^{\circ}$,then the length of the vector $\left|\frac{a}{2}-\frac{b}{3}\right|$ is

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\sqrt{3} \vec{c}=\overrightarrow{0}$,then the angle between $\vec{a}$ and $\vec{b}$ is

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{3}$. If a parallelogram is constructed with adjacent sides $2\vec{a}+3\vec{b}$ and $\vec{a}-\vec{b}$,then its shorter diagonal is of length

$M$ and $N$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$ respectively,then $\overline{AM} + \overline{AN} =$

If $P=(0,1,2)$,$Q=(4,-2,1)$,and $O=(0,0,0)$,then $\angle POQ$ is equal to