$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

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

If $m_1, m_2, m_3$ and $m_4$ are respectively the magnitudes of the vectors $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$, then the correct order of $m_1, m_2, m_3$ and $m_4$ is

$A$ vector $\bar{a}$ has components $1$ and $2p$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\bar{a}$ has components $1$ and $(p+1)$,then:

If $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ and $\overrightarrow{BC} = 6\hat{i} - 2\hat{j} + 3\hat{k}$ are the vectors along two sides of a triangle $ABC$,then the perimeter of triangle $ABC$ is:

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$. Then the triplet $(\alpha, \beta, \gamma)$ is