If $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + \hat{j} - \hat{k}$,and $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{p} = 2\vec{a} + \vec{b}$ and $\vec{q} = \vec{a} + 2\vec{b}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$,then $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}$ is:

$A$ zero vector has

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 $2 \alpha+\beta+\gamma$.

If $\hat{i}+4 \hat{j}+3 \hat{k}$,$\hat{i}+2 \hat{j}+3 \hat{k}$,and $3 \hat{i}+2 \hat{j}+\hat{k}$ are position vectors of $A$,$B$,and $C$ respectively,and if $D$ and $E$ are midpoints of sides $BC$ and $AC$,then $\overrightarrow{DE}$ is equal to:

$a$ and $b$ are two non-collinear vectors,then $xa + yb$ (where $x$ and $y$ are scalars) represents a vector which is