If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is:

Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$, $\vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-4 \hat{j}-12 \hat{k}$ be three vectors. If $\vec{p}$ is the projection of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection of $\vec{c}$ on $\vec{a}$, then $13 \vec{p}=$ (in $\vec{q}$)

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.

If $\overrightarrow{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{B} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{C} = 3\hat{i} + \hat{j}$,then the value of $t$ such that $\overrightarrow{A} + t\overrightarrow{B}$ is at a right angle to vector $3\hat{i} + 4\hat{j}$ is

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2\vec{a}+k\vec{b}+k\vec{c}|=3$, the positive value of $k$ is: