If the shortest distance between the lines (x - λ)/(- 2) = (y -...

Given: Equations of line,

$\dfrac{x - \lambda}{- 2} = \dfrac{y - 2}{1} = \dfrac{z - 1}{1}$ and $\dfrac{x - \sqrt{3}}{1} = \dfrac{y - 1}{- 2} = \dfrac{z - 2}{1}$

Now, we know that shortest distance between two lines is given by,

$\text{S.D} = \left|\dfrac{\left(\overset{\to}{a}2 - \overset{\to}{a}1\right) \cdot \overset{\to}{{b}_{1}} \times \overset{\to}{{b}_{2}}}{\left|\overset{\to}{{b}_{1}} \times \overset{\to}{{b}_{2}}\right|}\right|$

⇒ 1 = \left|\dfrac{\left|\begin{cases}\lambda - \sqrt{3} & 1 & - 1 \\ - 2 & 1 & 1 \\ 1 & - 2 & 1\end{cases}\right|}{\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 2 & 1 & 1 \\ 1 & - 2 & 1\end{cases}\right|}\right|

⇒ 1 = \left|\dfrac{\left(\lambda - \sqrt{3}\right)3 + 3 - 3}{\left|3\overset{^}{i} + 3\overset{^}{j} + 3\overset{^}{k}\right|}\right|

$⇒ \left|\left(\lambda - \sqrt{3}\right)\right| = \sqrt{3}$

$⇒ \lambda = 0, 2\sqrt{3}$

Hence, the sum is $2\sqrt{3}$