The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}$ is

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

$A$ vector $\vec{a}$ of length $2$ units makes an angle $60^{\circ}$ with each of the $X$-axis and $Y$-axis. If another vector $\vec{b}$ of length $\sqrt{2}$ units makes an angle $45^{\circ}$ with each of the $Y$-axis and $Z$-axis,then $\vec{a} \times \vec{b} = $

For a triangle $ABC$,let $\vec{p}=\vec{BC}$,$\vec{q}=\vec{CA}$ and $\vec{r}=\vec{BA}$. If $|\vec{p}|=2\sqrt{3}$,$|\vec{q}|=2$ and $\cos \theta = \frac{1}{\sqrt{3}}$ where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$,then $|\vec{p} \times (\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to:

If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be:

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 $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$