Find the equation of the plane equidistant from the points (2,8,3) and (4,12,9)

Answer 1

The plane must pass through the midpoint of the given points which is
(3,10,6)
A direction vector through the 2 points is
[2, 4, 6] or reduced to [1,2,3]
This must be a normal to the plane, thus the
equation of the plane would be

x + 2y + 3z = c
but (3,10,6) must lie on it, so
3 + 20 + 18 = c
c = 41

x + 2y + 3z = 41