The Solution is (2).
If $\overrightarrow{\mathrm{L}}=$ constant then $\vec{\tau}=0$
So $\vec{r} \times \vec{F}=0 \Rightarrow \vec{F}$ should be parallel to $\vec{r}$ so coefficient should be in same ratio. so $\frac{\alpha}{2}=\frac{3}{-6}=\frac{6}{-12}$
So $\alpha=-1$