The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

If $\hat{a}$ is a unit vector such that $(\bar{x}-\hat{a}) \cdot (\bar{x}+\hat{a}) = 8$,then $|\bar{x}| = $

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then $\cos ^2 A$ is equal to

The scalars $l$ and $m$ such that $la + mb = c,$ where $a, b$ and $c$ are given vectors,are equal to

If $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,then the least value of $|\hat{a}+\hat{b}|^2+|\hat{b}+\hat{c}|^2+|\hat{c}+\hat{a}|^2$ will be-

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is: