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

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:

Let $ABCD$ be a quadrilateral with $\overline{AB}=\bar{a}$,$\overline{AD}=\bar{b}$ and $\overline{AC}=3\bar{a}+2\bar{b}$. If its area is $\alpha$ times the area of the parallelogram with $AB$ and $AD$ as adjacent sides,then the value of $\alpha$ is equal to

The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

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

Two adjacent sides of a parallelogram are given by vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find the area of the parallelogram in square units.