Question

Find an equation of the plane whose points are equidistant from (8,-5,3) and (9,5,9).

Answer 1

The plane would have to pass through the midpoint of the line joining the
two given points and be perpendicular to that line

midpoint = ( (8+9)/2 , (-5+5)/2, (3+9)/2) = (17/2 , 0, 6)
The direction vector, which will become the normal to the plane
= < 9-8, 5+5, 9-3> = <1,10,6>

so the equation of our needed plane is
x + 10y + 6z = c
but (17/2, 0, 6) lies on it, so
17/2 + 0 + 36 = c
17 + 72 = 2c
c = 89/2

x + 10y + 6z = 89/2 is one form of your equation of the plane
another is
2x + 20y + 12z = 89