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

Find a unit vector perpendicular to both vectors $3i + 2j - k$ and $12i + 5j - 5k$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

If $a = i + j + k$,$b = i + 3j + 5k$,and $c = 7i + 9j + 11k$,then the area of the parallelogram having diagonals $a + b$ and $b + c$ is

If $A(3,2,-1), B(-2,2,-3)$ and $D(-2,5,-4)$ are the vertices of a parallelogram,then the area of the parallelogram is

If $\vec{a}+\vec{b}+\vec{c}=0,$ show that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .$ Interpret the result geometrically.