The adjacent sides of a parallelogram are $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find its area.

If $\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\bar{c}=\hat{i}+\hat{j}-2 \hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is:

If the area of a parallelogram,whose diagonals are $\vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \alpha\hat{k}$,is $\frac{\sqrt{93}}{2}$ sq. unit,then $\alpha = $

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,find $\vec{a} \times \vec{b}$.

$A$ vector perpendicular to both of the vectors $i + j + k$ and $i + j$ is