Let the shortest distance between the lines L: (x - 5)/(- 2) = (y -...

Given equations of lines are

$L: \dfrac{x - 5}{- 2} = \dfrac{y - \lambda}{0} = \dfrac{z + \lambda}{1} = k$

${L}_{1}: \dfrac{x + 1}{1} = \dfrac{y - 1}{1} = \dfrac{z - 4}{- 1}$

Here,

\overset{\to}{{b}_{1}} \times \overset{\to}{{b}_{2}} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 2 & 0 & 1 \\ 1 & 1 & - 1\end{cases}\right|

⇒ \overset{\to}{{b}_{1}} \times \overset{\to}{{b}_{2}} = - \overset{^}{i} - \overset{^}{j} - 2\overset{^}{k}

And,

\overset{\to}{{a}_{2}} - \overset{\to}{{a}_{1}} = 6\overset{^}{i} + \left(\lambda - 1\right)\overset{^}{j} + \left(- \lambda - 4\right)\overset{^}{k}

$∵$ Shortest distance is $2\sqrt{6}$.

$⇒ 2\sqrt{6} = \left|\dfrac{- 6 - \lambda + 1 + 2\lambda + 8}{\sqrt{1 + 1 + 4}}\right|$

$⇒ | \lambda + 3 | = 12 ⇒ \lambda = 9, - 15$

$⇒ \alpha = - 2k + 5, \gamma = k - \lambda$ where $k \in R$

So,

$\alpha + 2\gamma = 5 - 2\lambda = - 13, 35$