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 $8 a\beta$  

The lines $\frac{x-1}{2}-\frac{2-y}{3}-\frac{z-3}{\alpha }$ and $\frac{x-4}{5}-\frac{y-1}{2}-\frac{y-1}{2}=\frac{z}{\beta }$  are inter sect

$\left|\begin{array}{l}{x}_2-x_2{y}_2-y_1{z}_2-z_1\\ a_{1 }{b}_1 c_1\\ a_2{b}_2{c}_2\end{array}\right|=0$

$\left|\begin{array}{l}4-11-20-3\\ 23\alpha \\ 52\beta \end{array}\right|=0$

$\left|\begin{array}{l}3-1-3\\ 2 3 \alpha \\ 52 \beta \end{array}\right|=0$

$3\left(3\beta -2\alpha \right)+1\left(2\beta -5\alpha \right)-3\left(4-15\right)=0$

$9\beta -6\alpha +2\beta -5\alpha +33=0$

$11\beta -11\alpha +33=0$

$\alpha -\beta =3\Rightarrow \alpha =3+\beta$

$8\alpha \beta =8\left(3+\beta \right)=8\left(3+{\beta }^2\right)$

$f^{,}\left(\beta \right)=8\left(3\beta +{\beta }^2\right)$

$f^{,}\left(\beta \right)=16$

$f^{,}\left(\beta \right)=0\Rightarrow \beta =\frac{3}{2}:{\left[f^{ }\left(\beta \right)\right]}_{\beta =\frac{3}{2}}=16>0$

$f\left(\beta \right)$ is minimum of $\beta =-\frac{3}{2}$

$\Rightarrow \alpha =3-\frac{3}{2}=\frac{3}{2}$

$\left[8\alpha \beta \right]=\left|8\alpha \frac{3}{2}\times -\frac{3}{2}\right|=\left|-18\right|=18$