If the shortest distance between the lines $\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$, then the sum of all possible values of $\lambda$ is

Given: Equations of line,

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

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

$\text{S.D}=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\overrightarrow{b_1}\times \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right|$

$\Rightarrow 1=\left|\frac{\left|\begin{array}{ccc}\lambda -\sqrt{3}& 1& -1\\ -2& 1& 1\\ 1& -2& 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& 1\\ 1& -2& 1\end{array}\right|}\right|$

$\Rightarrow 1=\left|\frac{\left(\lambda -\sqrt{3}\right)3+3-3}{\left|3\hat{i}+3\hat{j}+3\hat{k}\right|}\right|$

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

$\Rightarrow \lambda =0, 2\sqrt{3}$

Hence, the sum is $2\sqrt{3}$