We first find the plane perpendicular to these two lines using cross-product (note that the coefficients here are the directions of the two lines)

i j k

1 1 0

1 0 2

2i - 2j - k

Then, we determine the constant for each line.

For the first line, letting s = 0, we get (-1, 0, 3); substituting, we get 2 * -1 - 2 * 0 - 1 * 3 = -5

For the second line, letting t = 0, we get, substituting (2, 0, 0), 2 * 2 = 4

| 4 - -5|/√2^{2} + (-2)^{2} + (-1)^{2} = 9/√9 = 9/3 = 3.

Now, drop down a distance from 3 from the T plane in the direction (2, -2, -1); nicely, we can just use this as the step.

Thus, we have lines (s-1, s, 3) and (2+t, 0, 2t) - (2, -2, -1) = (t, 2, 2t + 1)

From setting the second coordinates equal, s = 2. Substituting into the first coordinate, 1 = t.

The lines meet at (1, 2, 3)

Then, projecting back to the T plane, (1, 2, 3) + (2, -2, -1) = (3, 0, 2)

As a check, note that, in the T line, when t = 1, we get (2+t, 0, 2t) = (2+1, 0, 2*1) = (3, 0, 2), which is the point

Thus, our points, again, are (1, 2, 3) on the s line and (3, 0, 2) on the t line.