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 $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $-\hat{i}+2 \hat{j}+\hat{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\overrightarrow{c} = 2\hat{i} + 3\hat{j} - \hat{k}$,and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i} + \hat{j} + \hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i} + \alpha\hat{j} + \hat{k}$,then the value of $\alpha$ is:

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

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

Let $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$. 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 $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is equal to