Hello again:

**note: z-coord is always 0.0 <<<< !!!** If you expect a 3D point you have to deal p1 and p2 as 2 vectors.

If you do the cross product "(Wedge)" of the 2 vectors (0,0,0) - p1 & (0,0,0) - p2, then you get the solution of a 3Dvector perpendicular to the plane are (0,0,0)-p1-p2.

The vector for this case is (749,875 - 417,559, -0.683505).

If you reduce the length to 1 (unit vector) then get (0.873681, -0.486499, -0.000796353)

You can draw the data and verify that the new vector is perpendicular a (0,0,0)-p1 & (0,0,0)-p2 (in their respectives planes ) .

I hope I have solved your questions on the normal vector and its Z coord.

Greetings.