Given $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$

and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$

angle between $$\vec{a}$$ and $$\vec{b}$$ is given by

$$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

We have, $$\cos \theta < 0(\because$$ angle between $$\vec{a}$$ and $$\vec{b}$$ is obtuse)

$$\begin{aligned} & \Rightarrow \quad \vec{a} \cdot \vec{b}<0 \\ & \Rightarrow \alpha t^2-12+6 \alpha t<0 \forall t \in \mathbb{R} \end{aligned}$$

If $$\alpha=0$$, then $$-12<0$$ (condition holds)

If $$\alpha \neq 0 \Rightarrow \alpha<0\quad \text{.... (i)}$$

And maximum value of $$\alpha t^2+6 \alpha t-12<0$$

$$\begin{aligned} & \Rightarrow \frac{-D}{4 a}<0 \text { (where } D \text { is discriminant and } a=\alpha) \\ & \Rightarrow \frac{36 \alpha^2+48 \alpha}{4 \alpha}>0 \\ & \Rightarrow \alpha>\frac{-4}{3} \\ & \therefore \alpha \in\left(\frac{-4}{3}, 0\right] \end{aligned}$$