If the shortest distance between the lines (x - 4)/(1) = (y +...

$\dfrac{x - 4}{1} = \dfrac{y + 1}{2} = \dfrac{z}{- 3}$

$\dfrac{x - \lambda}{2} = \dfrac{y + 1}{4} = \dfrac{z - 2}{- 5}$

The shortest distance between the lines
$= \left|\dfrac{( \overset{\to}{a} - \overset{\to}{b} ) \cdot \left(\overset{\to}{{d}_{1}} \times \overset{\to}{{d}_{2}}\right)}{\left|\overset{\to}{{d}_{1}} \times \overset{\to}{{d}_{2}}\right|}\right|$

$a ≡ \left(4, - 1, 0\right), b ≡ \left(\lambda, - 1, 2\right)$

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

⇒ \overset{\to}{a} - \overset{\to}{b} = \left(4 - \lambda\right)\overset{^}{i} - 2\overset{^}{k}

So, the shortest distance is given by,

D = \left|\dfrac{\left|\begin{cases}\lambda - 4 & 0 & 2 \\ 1 & 2 & - 3 \\ 2 & 4 & - 5\end{cases}\right|}{\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 2 & - 3 \\ 2 & 4 & - 5\end{cases}\right|}\right|

⇒ D = \left|\dfrac{( \lambda - 4 ) ( - 10 + 12 ) - 0 + 2 ( 4 - 4 )}{| 2\overset{^}{i} - 1\overset{^}{j} + 0\overset{^}{k} |}\right|

$⇒ \dfrac{6}{\sqrt{5}} = \left|\dfrac{2 ( \lambda - 4 )}{\sqrt{5}}\right|$

$⇒ 3 = | \lambda - 4 |$

$⇒ \lambda - 4 = \pm 3$

$⇒ \lambda = 7, 1$

So, sum of all possible values of $\lambda$ is $8$.