If $D, E$ and $F$ are respectively the mid-points of $AB, AC$ and $BC$ in $\triangle ABC$,then $\overrightarrow{BE} + \overrightarrow{AF}$ is equal to :

$A$ point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors) divides the line segment joining $A$ and $B$ in the ratio $2:1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$,then find the position vector of $B$.

In the figure,a vector $x$ satisfies the equation $x - w = v$. Then $x =$

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}$ be two vectors such that $\vec{a} \cdot \vec{b}=1$,$\cos(\theta) = \frac{1}{3}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,and the components of $\vec{b}$ with respect to $(\hat{i}, \hat{j}, \hat{k})$ are integers. Then the number of possible vectors that represent $\vec{b}$ is

If $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$ and $a \cdot b = 27$,then $|a| =$