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

Find the unit vector in the direction of the sum of vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{j} + \hat{k}$.

If $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$,then under what condition are the three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ collinear?

Let $\overline{i}-2 \overline{j}+\overline{k}, \overline{i}+\overline{j}-2 \overline{k}, 2 \overline{i}-\overline{j}-\overline{k}$ and $\overline{i}+\overline{j}+\overline{k}$ be the position vectors of four points $A, B, C$ and $D$ respectively. If a point $P$ divides $AB$ in the ratio $2:1$ internally and a point $Q$ divides $CD$ in the ratio $1:2$ externally,then the ratio in which the point with position vector $5 \overline{i}-6 \overline{j}-5 \overline{k}$ divides $PQ$ is

If $a$ and $b$ are the position vectors of $A$ and $B$ respectively,then the position vector of a point $C$ on $AB$ produced such that $\overrightarrow{AC} = 3\overrightarrow{AB}$ is

The diagonals of a parallelogram are the vectors $\vec{d_1} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$ and $\vec{d_2} = -\hat{i} - 2 \hat{j} - 8 \hat{k}$. Then the length of the shorter side of the parallelogram is