$L$ and $M$ are two points with position vectors $2 \vec{a}-\vec{b}$ and $\vec{a}+2 \vec{b}$ respectively. The position vector of the point $N$ which divides the line segment $LM$ in the ratio $2:1$ externally is

If $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$,$b = 2 \hat{i} + 3 \hat{j} + \hat{k}$,$c = 8 \hat{i} + 13 \hat{j} + 9 \hat{k}$ and $x a + y b + z c = 0$,then $\frac{x y}{z^2} =$

If $\bar{x}$ is a non-zero vector and $k < 0, k \neq -1$,then $\frac{k \bar{x}}{|\bar{x}|}$ is $.........$

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$

If the vector $\bar{i}-7 \bar{j}+2 \bar{k}$ is along the internal bisector of the angle between the vectors $\bar{a}$ and $-2 \bar{i}-\bar{j}+2 \bar{k}$ and the unit vector along $\bar{a}$ is $x \bar{i}+y \bar{j}+z \bar{k}$ then $x=$