For what value of a, $\overrightarrow{A}=2\hat{i}+a\hat{j}+\hat{k}$  will be perpendicular to $\overrightarrow{B}=4\hat{i}-2\hat{j}-2\hat{k}$ ?

$\overrightarrow{A}\perp \overrightarrow{B}$ , if $\overrightarrow{A}.\overrightarrow{B}=AB\cos {90}^0=0$

$\left(2\hat{i}+a\hat{j}+\hat{k}\right)\left(4\hat{i}-2\hat{j}-2\hat{k}\right)=0$

 or                               $8-2a-2=0$  or $a=3$