$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

$A$ vector whose modulus is $\sqrt{51}$ and makes the same angle with $a = \frac{i - 2j + 2k}{3}$,$b = \frac{-4i - 3k}{5}$,and $c = j$ is:

If the position vectors of three points $A, B, C$ are $\hat{i}+2\hat{j}+\hat{k}$, $2\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ respectively, then the perpendicular distance of the point $C$ from the line $AB$ is

Let $P$ be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\overrightarrow{AP}|=\frac{\sqrt{5}}{2}$, then the area of the triangle $ABP$ is:

If in a $\Delta ABC$,$O$ and $O^{\prime}$ are the incentre and orthocentre respectively,then $\vec{O^{\prime}A} + \vec{O^{\prime}B} + \vec{O^{\prime}C}$ is equal to

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{2}$ and $\vec{a} \cdot \vec{b}=-1$,then the angle between $\vec{a}$ and $\vec{b}$ is