The vector $c$ directed along the internal bisector of the angle between the vectors $a = 7i - 4j - 4k$ and $b = -2i - j + 2k$ with $|c| = 5\sqrt{6}$ is

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$,$|\bar{a}|=3$,$|\bar{b}|=4$,and $|\bar{c}|=5$. Then,find the value of $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$.

Let $\sqrt{3} \hat{i} + \hat{j}$,$\hat{i} + \sqrt{3} \hat{j}$ and $\beta \hat{i} + (1 + \beta) \hat{j}$ respectively be the position vectors of the points $A, B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $OA$ and $OB$ is $\frac{3}{\sqrt{2}}$,then the sum of all possible values of $\beta$ is:

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{C}$. Then $\overrightarrow{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

Assertion $(A)$: The scalar product of force $\vec{F}$ and displacement $\vec{r}$ is equal to the work done.
Reason $(R)$: The work done is not a scalar quantity.

Let the points $A, B$ and $P$ be $(-2, 2, 4), (2, 6, 3)$ and $(1, 2, 1)$ respectively. The magnitude of the moment of the force represented by $\overrightarrow{AB}$ and acting at $A$ about $P$ is