Given $2\overline{i}+\lambda \overline{j}-\overline{k}$ and $4\overline{i}-2\overline{j}+2\overline{k}$ are perpendicular

$\therefore (2\overline{i}+\lambda \overline{j}-\overline{k})\cdot (4\overline{i}-2\overline{j}+2\overline{k})=0$

$\Rightarrow \left(2\right)\left(4\right)+\left(\lambda \right)(-2)-\left(1\right)\left(2\right)=0$

$\Rightarrow 8-2\lambda -2=0\Rightarrow 2\lambda =6\Rightarrow \boxed{\lambda =3}$