If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ sq. units,then the area of the parallelogram having $3 \vec{a}+\vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides,in square units,is

If $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$,then the value of $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times 2\vec{b})]$ is:

Find the unit vector perpendicular to both vectors $\vec{a}$ and $\vec{b}$.

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

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

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ is