One side and one diagonal of a parallelogram are represented by $3 \hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ respectively. Then,the area of the parallelogram in square units is:

If a non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ and the plane determined by the vectors $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$,then the angle between the vectors $\vec{a}$ and $\hat{i}+\hat{j}+\hat{k}$ is

Let $\overline{a}, \overline{b}, \overline{c}$ be three vectors such that $|\overline{a}|=\sqrt{3}$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$ and the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$. If $\overline{a}$ is perpendicular to the vector $\overline{b} \times \overline{c}$,then $|\overline{a} \times(\overline{b} \times \overline{c})|$ is equal to

$A$ unit vector which is perpendicular to $i + 2j - 2k$ and $-i + 2j + 2k$ 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 parallelogram whose adjacent sides are $\hat{i}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ is