We know that, $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$

$\Rightarrow \cos \theta =\frac{(\alpha \hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}})·(\alpha \hat{\mathrm{i}}+2\alpha \hat{\mathrm{j}}-2\hat{\mathrm{k}})}{\sqrt{{\alpha }^2+4+4}\sqrt{{\alpha }^2+4{\alpha }^2+4}}$

$\Rightarrow \cos \theta =\frac{{\alpha }^2-4\alpha -4}{\sqrt{{\alpha }^2+8}\sqrt{5{\alpha }^2+4}}$

Since $\theta$ is acute,

$\Rightarrow {\alpha }^2-4\alpha -4>0$

$\Rightarrow {\alpha }^2-4\alpha +4>8$

$\Rightarrow (\alpha -2{)}^2>8$

$\Rightarrow \alpha -2>2\sqrt{2}$ or $\alpha -2<-2\sqrt{2}$

$\alpha >2+2\sqrt{2}$ or $\alpha <2-2\sqrt{2}$

So, the least positive integral value of $\alpha$ is $5$.