The set of all $$\alpha$$, for which the vectors $$\vec{a} = \alpha t\hat{i} + 6\hat{j} - 3\hat{k}$$ and $$\vec{b} = t\hat{i} - 2\hat{j} - 2\alpha t\hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is

For an obtuse angle, the dot product $$\vec{a} \cdot \vec{b} < 0$$.

$$(\alpha t)(t) + (6)(-2) + (-3)(-2\alpha t) < 0$$

$$\alpha t^2 + 6\alpha t - 12 < 0$$.

For a quadratic $$At^2 + Bt + C$$ to be always negative:

$$A < 0 \implies \mathbf{\alpha < 0}$$.

Discriminant $$D < 0 \implies (6\alpha)^2 - 4(\alpha)(-12) < 0$$.

$$36\alpha^2 + 48\alpha < 0 \implies 12\alpha(3\alpha + 4) < 0$$.

The roots are $$0$$ and $$-4/3$$. The inequality holds for $$\alpha \in (-4/3, 0)$$.

Checking $$\alpha = 0$$: The equation becomes $$-12 < 0$$, which is true. So, $$\alpha \in \mathbf{(-4/3, 0]}$$. (Option C)