If $a, b, c$ are the position vectors of three collinear points,then the existence of scalars $x, y, z$ (not all zero) is such that:

The angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of the $X$-axis is:

If $PQRST$ is a pentagon,then the resultant of forces $\overline{PQ}, \overline{PT}, \overline{QR}, \overline{SR}, \overline{TS}$ and $\overline{PS}$ is

If the vectors $\vec{AB} = -3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$,then the length of the median through $A$ is

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

$A$ 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 $B$.