If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\frac{\pi}{6}$,then $|a \times b|^2$ is equal to

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$.

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$

If $\bar{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\bar{b}=\hat{i}-2 \hat{j}-2 \hat{k}$,$\bar{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$ and if $\bar{d}$ is a vector perpendicular to both $\bar{b}$ and $\bar{c}$,and $\bar{a} \cdot \bar{d}=18$,then $|\bar{a} \times \bar{d}|^2=$

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is