Find the equation of the plane in R^3 that is perpendicular to (1,2,3) and contains the point (3,2,1).
So the normal to the plane is (1,2,3)
So: x+2y+3z=d
Sub the point in and solve for d.
(3) + 2(2) + 3(1) + d = 0
10 + d = 0
d = -10
So the equation is
x + 2y + 3z -10 = 0
Is this the correct method?