The work done in moving an object along the vector $\vec{d} = 3i + 2j - 5k$,if the applied force is $\vec{F} = 2i - j - k$,is

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}$,$\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}$,and $\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ be three vectors such that the projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{a}$. If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$,then $|\vec{b}|$ is equal to

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

Let $\bar{a} = 4\bar{i} + 3\bar{j}$ and $\bar{b}$ be two perpendicular vectors in the $XOY$-plane. $A$ vector $\bar{c}$ in the same plane and having projections $1$ and $2$ respectively on $\bar{a}$ and $\bar{b}$ is

If $a = i - j$ and $b = i + k$,then a unit vector coplanar with $a$ and $b$ and perpendicular to $a$ is