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 _______.

Two vectors $$\alpha\hat{i} - 2\hat{j} + 2\hat{k}$$ and $$\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$$ are given such that the angle between them is an acute angle. 

When we have two vectors which have an acute angle between them, the dot product of the vectors will be positive. 

$$\left(\alpha\times\alpha\right)+\left(-2\times2\alpha\right)+\left(2\times-2\right)>0$$

$$\alpha^2-4\alpha-4>0$$

For $$\alpha=1,\ \alpha^2-4\alpha-4=-7$$

For $$\alpha=2,\ \alpha^2-4\alpha-4=-8$$

For $$\alpha=3,\ \alpha^2-4\alpha-4=-7$$

For $$\alpha=4,\ \alpha^2-4\alpha-4=-4$$

For $$\alpha=5,\ \alpha^2-4\alpha-4=1$$

Hence, for $$\alpha=5$$, the dot product is positive for the first time. 

Hence, the minimum value of $$\alpha$$ for which the angle between two vectors $$\alpha\hat{i} - 2\hat{j} + 2\hat{k}$$ and $$\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$$ is acute is 5. 

$$\therefore\ $$ The required answer is 5.