Find equation of plane perpendicular to (1, 2, 3) passing through (4, 5, 6):
- x+2y+3z = 32 (= 4 + 10 + 18)
- x = 4
- x+2y+3z = 0
- x+y+z = 15
Show answer and solution
Correct answer: A. x+2y+3z = 32 (= 4 + 10 + 18)
Normal n = (1,2,3).
**Plane:** n · (r - p) = 0
We have → 1(x-4) + 2(y-5) + 3(z-6) = 0 → x + 2y + 3z.
So → 1(x-4) + 2(y-5) + 3(z-6) = 32.