If $a, b, c$ are the position vectors of the vertices $A, B, C$ of the triangle $ABC,$ then the centroid of $\Delta ABC$ is

Let $ABCDEF$ be a regular hexagon with the vertices $A, B, C, D, E, F$ in counterclockwise order. Then the vector $\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF}$ is equal to

$A$ zero vector has

The vertices of triangle $ABC$ are $A \equiv (3, 0, 0)$,$B \equiv (0, 0, 4)$,and $C \equiv (0, 5, 4)$. Find the position vector of the point $D$ where the angle bisector of $\angle A$ meets $BC$.

$ABCDEF$ is a regular hexagon whose centre is $O$. Then, $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$ is equal to (in $\vec{AO}$)

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$