Given, lines

$$ \begin{aligned} & \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\ &\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i) \end{aligned} $$

Any point on the line (i)

$$ x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3 $$

and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu$ ............(ii)

Any point on line (ii)

$$ \Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu $$

Since, given lines intersects

$$ \begin{aligned} & \therefore 2 \lambda+1=5 \mu+4 ..........(iii)\\\\ & 3 \lambda+2=2 \mu+1 ............(iv)\\\\ & \text { and } \alpha \lambda+3=\beta \mu ..........(iv) \end{aligned} $$

On solving (iii) and (iv), we get

$$ \lambda=-1, \mu=-1 $$

On putting value of $\lambda$ and $\mu$ in (v), we get

$$ \begin{array}{cc} & \alpha(-1)+3=-\beta \\\\ &\Rightarrow \alpha=\beta+3 \end{array} $$

Now,

$$ \begin{aligned} & 8 \alpha \beta=8 (\beta+3)(\beta) \\\\ &= 8\left(\beta^2+3 \beta\right) \\\\ & =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\\\ & =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\} \end{aligned} $$

$$=8\left(\beta+\frac{3}{2}\right)^2-18$$

Here, minimum value $=-18$

$\therefore$ Magnitude of the minimum value of $8 \alpha \beta$ is 18 .