The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ is acute, is ___________.

<p>$$\begin{aligned} & \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\ & \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\ & \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 \\ & \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} \\ & \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2<2 \sqrt{2} \\ & \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} \\ & \alpha \in(-\infty,-0.82) \cup(4.82, \infty) \end{aligned}$$</p> <p>Least positive integral value of $\alpha \Rightarrow 5$</p>