If the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x}{1}=\dfrac{y}{\alpha}=\dfrac{z-5}{1}$ is $\dfrac{5}{\sqrt6}$, then the sum of all possible values of $\alpha$ is