If the lines (x - 1)/(2) = (2 - y)/(- 3) = (z - 3)/(α) and (x -...

Let,

${L}_{1} = \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{\alpha} = \lambda$

${L}_{2} = \dfrac{x - 4}{5} = \dfrac{y - 1}{2} = \dfrac{z - 0}{\beta} = \mu$

For point of intersection

$2\lambda + 1 = 5\mu + 4 . . . \left(i\right)</p><p>$3\lambda + 2 = 2\mu + 1 . . . \left(ii\right)$</p><p>$\alpha\lambda + 3 = \beta\mu + 0 . . . \left(iii\right)

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

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

Let $y = 8\alpha\beta$

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

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

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

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