If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is:

$A$ unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is . . . . . . where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three unit vectors,out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\vec{a}$ makes with vectors $\vec{b}$ and $\vec{c}$ respectively and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b},$ then $|\alpha - \beta|$ is equal to .............. $^o$

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$, $\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2, \bar{b} \cdot \bar{c}=8$ and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then the value of $|\bar{a} \times(\bar{b} \times \bar{c})|$ is