3d Geometry

Let the point \((-1, \alpha, \beta)\) lie on the line of the shortest distance between the lines \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}\) and \(\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}\). Then \((\alpha-\beta)^2\) is equal to _________.