If $a, b$ and $c$ are three vectors such that $|a|=|b|=2$,$a \cdot b=2$ and $a+b+c=0$,then $|c|$ is equal to

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and the points represented by position vectors $\bar{a}-2 \bar{b}+3 \bar{c}$, $-4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $x \bar{a}-9 \bar{b}+z \bar{c}$ are collinear, then $2x-z=$

$a, b, c$ are non-coplanar vectors. If $a+3 b+4 c=x(a-2 b+3 c)+y(a+5 b-2 c)+z(6 a+14 b+4 c)$,then $x+y+z=$

The direction cosines of the vector $\hat{i}-2\hat{j}+3\hat{k}$ are . . . . . . .

Given $\overrightarrow{p} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$,$\overrightarrow{a} = \hat{i} + \hat{j}$,$\overrightarrow{b} = \hat{j} + \hat{k}$,$\overrightarrow{c} = \hat{i} + \hat{k}$ and $\overrightarrow{p} = x \overrightarrow{a} + y \overrightarrow{b} + z \overrightarrow{c}$,then $x, y, z$ are respectively: