Then don't use that method! Instead, use the "long way" SpringFan25 mentions:

Also, "r= ai+bj+ck + s(xi+yj+zk)" is an unfortunate way to write the line because the x, y, and z in that are simply coefficients, NOT the (x, y, z) coordinates of a point. I am going to write instead "r= xi+yj+ zk= ai+bj+ ck+ s(Ai+ Bj+ Ck)" so that (x, y, z)= (a+ sA, b+ sB, c+ sC).

Now you know that Ai+ Bj+ Ck is a vector pointing in the direction of the line. A **plane**perpendicular to the line is given by \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\) where \(\displaystyle (x_0, y_0, z_0)[/itex] is any point on the line (take s to be any value- 0 for example).

Now, the point on the second line closest to the first line is where that second line crosses that plane. Put the (x, y, z) values given by the equation of the second line, in terms of some parameter, t, into the equation of the plane to find t and then the point.

Finally, find the distance between those two points.\)