$A, B, C, D, E$ are five coplanar points,then $\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} + \overrightarrow{AE} + \overrightarrow{BE} + \overrightarrow{CE}$ is equal to

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to:

If $a, b, c$ are non-collinear vectors such that for some scalars $x, y, z,$ $xa + yb + zc = 0,$ then

Let $\vec{i}+\vec{j}+\vec{k}$, $a_1 \vec{i}+b_1 \vec{j}+c_1 \vec{k}$, $a_2 \vec{i}+b_2 \vec{j}+c_2 \vec{k}$, and $a_3 \vec{i}+b_3 \vec{j}+c_3 \vec{k}$ be the position vectors of the points $A, B, C, D$ respectively. The position vector of the centroid of the triangular face $BCD$ is $\frac{2}{3}(\vec{i}+\vec{j}+\vec{k})$. If $\alpha \vec{i}+\beta \vec{j}+\gamma \vec{k}$ is the position vector of the centroid of the tetrahedron $ABCD$, then find the value of $2 \alpha+\beta+\gamma$.

The position vector of a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

If $P$ divides the line segment joining the points $A$ and $B$ in the ratio $2:1$ and the position vectors of $A$ and $B$ are $\hat{i}-2\hat{j}$ and $-3\hat{i}+5\hat{j}$ respectively,then the position vector of $P$ is