The line connected the points where the lines are closest together must be perpendicular to both and so must lie in a plane that is perpendicular to both lines. Any plane perpendicular to the first line must be of the form x+ y- z= A for some constant A. Any plane perpendicular to the second line must be of the form x+ 2z= B for some constant B. Subtracting those two equations eliminates x and gives y- 3z= A- B, a constant, for all x and y in the plane. In particular, setting y= 1+ s and z= -1- s, as in the first line, gives 1+ s+ 3+ 3x= 4s+ 4= A- B. Setting y= 2, z= 2t as in the second line, gives 2- 6t= A- B. Putting those together, 4s+ 4= 2- 6t so that s= -(3t+ 1)/2.