If the shortest between the lines ${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$ and ${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$ is 6, then the square of sum of all possible values of $\lambda$ is :

Shortest distance between the lines<br/><br/> $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$<br/><br/> $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6<br/><br/> Vector along line of shortest distance<br/><br/> $=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k}$ (its magnitude is $\sqrt{6}$ )<br/><br/> Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=\pm 6$ <br/><br/>$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$<br/><br/> So, square of sum of these values <br/><br/>$=(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384$