The area of a parallelogram whose adjacent sides are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$ is . . . . . . sq. units.

If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is:

$A$ unit vector perpendicular to each of the vectors $2i - j + k$ and $3i + 4j - k$ is equal to

Let $\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be two vectors. If $\vec{a}$ is a vector such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then $|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=$

If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\pi/6$,then find $|a \times b|^{2}$.

The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ is: