Let,

$L_1=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\alpha }=\lambda$

$L_2=\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{\beta }=\mu$

For point of intersection

$2\lambda +1=5\mu +4 ... \left(\mathrm{i}\right)$

$3\lambda +2=2\mu +1 ... \left(\mathrm{ii}\right)$

$\alpha \lambda +3=\beta \mu +0 ... \left(\mathrm{iii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get $\lambda =\mu =-1$

Now, from $\left(\mathrm{iii}\right)$, we get $\alpha -\beta =3$

Let $y=8\alpha \beta$

$\Rightarrow y=8\alpha \left(\alpha -3\right)$

$\Rightarrow y=8\left({\alpha }^2-3\alpha +\frac{9}{4}\right)-18$

$\Rightarrow y=8{\left(\alpha -\frac{3}{2}\right)}^2-18$

So, Minimum value of $y=-18$ at $\alpha =+\frac{3}{2}$