If the lines $\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}$ and $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$ intersect, then the magnitude of the minimum value of $8 \alpha \beta$ is _____________.

Given, lines <br/><br/>$$ \begin{aligned} & \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\ &\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i) \end{aligned} $$ <br/><br/>Any point on the line (i) <br/><br/>$x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3$ <br/><br/>and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu$ ............(ii) <br/><br/>Any point on line (ii) <br/><br/>$\Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu$ <br/><br/>Since, given lines intersects <br/><br/>$$ \begin{aligned} & \therefore 2 \lambda+1=5 \mu+4 ..........(iii)\\\\ & 3 \lambda+2=2 \mu+1 ............(iv)\\\\ & \text { and } \alpha \lambda+3=\beta \mu ..........(iv) \end{aligned} $$ <br/><br/>On solving (iii) and (iv), we get <br/><br/>$\lambda=-1, \mu=-1$ <br/><br/>On putting value of $\lambda$ and $\mu$ in (v), we get <br/><br/>$$ \begin{array}{cc} & \alpha(-1)+3=-\beta \\\\ &\Rightarrow \alpha=\beta+3 \end{array} $$ <br/><br/>Now, <br/><br/>$$ \begin{aligned} & 8 \alpha \beta=8 (\beta+3)(\beta) \\\\ &= 8\left(\beta^2+3 \beta\right) \\\\ & =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\\\ & =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\} \end{aligned} $$ <br/><br/>$=8\left(\beta+\frac{3}{2}\right)^2-18$ <br/><br/>Here, minimum value $=-18$ <br/><br/>$\therefore$ Magnitude of the minimum value of $8 \alpha \beta$ is 18 .