If the vertices of a $\triangle ABC$ are $A=(2,3,5)$,$B=(-1,3,2)$,and $C=(3,5,-2)$,then the area of the $\triangle ABC$ (in sq. units) is

If $a \times b = b \times c \ne 0,$ where $a, b$ and $c$ are coplanar vectors,then for some scalar $k$

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$,where $\alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$,then $\alpha$ is equal to:

If $|\bar{u}| = 8$ and $|\bar{v}| = 12$ with an angle of $150^{\circ}$ between them,then find $|\bar{u} \times \bar{v}|$.

The unit vector perpendicular to the vectors $i - j + k$ and $2i + 3j - k$ is

If $\overrightarrow{a}+2 \overrightarrow{b}+3 \overrightarrow{c}=\overrightarrow{0}$,then $\overrightarrow{a} \times \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}$ is equal to