If $\overrightarrow{A}=\hat{i}+2\hat{k}-\hat{k},\overrightarrow{B}=-\hat{i}+\hat{j}-2\hat{k}$  then angle between $\overrightarrow{A}$  and $\overrightarrow{B}$  is

$\overrightarrow{A}=\hat{i}+2\hat{j}-\hat{k};\overrightarrow{B}=\hat{i}+\hat{j}-2\hat{k}$

$A=\left|\overrightarrow{A}\right|=\sqrt{1^2+2^2{\left(-1\right)}^2}=\sqrt{6}$

$B=\left|\overrightarrow{B}\right|=\sqrt{{\left(-1\right)}^2+1^2{\left(-2\right)}^2}=\sqrt{6}$

$\overrightarrow{A}.\overrightarrow{B}=\left(\hat{i}+2\hat{j}-\hat{k}\right).\left(-\hat{i}+\hat{j}-2\hat{k}\right)$

        $=-1+2+2=3$

If $\theta$ is angle between $\overrightarrow{A}$  and $\overrightarrow{B}$ ,

 then $\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$

               $3=\sqrt{6}\sqrt{6}\cos \theta$

$\cos \theta =\frac{3}{6}=\frac{1}{2}$  or $\theta =\frac{\pi }{3}$