Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then:

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:

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is

If $\overrightarrow{OA}=2 \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB}=3 \hat{i}-\hat{k}$ and $\overrightarrow{OC}=2 \hat{j}+3 \hat{k}$ are the position vectors of the points $A, B$ and $C$,then a unit vector perpendicular to the plane containing $A, B$ and $C$ is

$A$ vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$,and the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2\hat{j}+2\hat{k}$ is