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

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c}=|\bar{c}|$,$|\bar{c}-\bar{a}|=2 \sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$. Then $|(\bar{a} \times \bar{b}) \times \bar{c}|=$

The area of the quadrilateral $ABCD$ with vertices $A(2, 1, 1)$,$B(1, 2, 5)$,$C(-2, -3, 5)$,and $D(1, -6, -7)$ is equal to

The vectors $a = xi + yj + zk$ and $b = j$ are such that $a, c, b$ form a right-handed system. Then $c$ 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.

If $\vec{x}$ is a unit vector such that $\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = -\hat{i} + \hat{k}$,then $\vec{x}$ is: