1. Let $\vec{a}=\alpha \hat{i}-2\hat{j}+2\hat{k}, \qquad \vec{b}=\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}.$

2. For the angle between two vectors to be acute, their dot product must be positive: $\vec{a}\cdot\vec{b}>0.$

3. Compute the dot product: $\vec{a}\cdot\vec{b}=(\alpha)(\alpha)+(-2)(2\alpha)+(2)(-2).$ $=\alpha^2-4\alpha-4.$

4. So we need $\alpha^2-4\alpha-4>0.$

5. Solve the corresponding quadratic equation: $\alpha^2-4\alpha-4=0.$ Using the quadratic formula, $\alpha=\frac{4\pm\sqrt{16+16}}{2}=\frac{4\pm\sqrt{32}}{2}=2\pm2\sqrt{2}.$

6. Since the parabola opens upward, $\alpha^2-4\alpha-4>0$ for $\alpha<2-2\sqrt{2} \quad \text{or} \quad \alpha>2+2\sqrt{2}.$

7. We need the least positive integer value of $\alpha$ satisfying this. Now, $2+2\sqrt{2} \approx 2+2(1.414)=4.828.$ Hence the least positive integer greater than this is $\alpha=5.$

8. Verification: For $\alpha=5$, $\vec{a}\cdot\vec{b}=25-20-4=1>0,$ so the angle is indeed acute.

Therefore, the least positive integral value is $\boxed{5}.$