If $\theta$ is the angle between two unit vectors $a$ and $b$,then $\sin(\theta/2) = $ ......

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:

If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and each one of $\vec{a}, \vec{b}, \vec{c}$ is perpendicular to the sum of the remaining two,then find $|\vec{a}+\vec{b}+\vec{c}|$.

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,$\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$

Forces of magnitudes $3$ and $2$ units acting in the directions $5\hat{i} + 3\hat{j} + 4\hat{k}$ and $3\hat{i} + 4\hat{j} - 5\hat{k}$ respectively act on a particle which is displaced from the points $(1, -1, -1)$ to $(3, 3, 1)$. The work done by the forces is equal to