Equation Of A Line In Space

The equation of plane through the point (2,2,1) is
a(x-2) + b(y-2) + c(z-1) = 0 ………(1)
Since this line passes through (9,3,6)
a(9-2) + b(3-2) + c(6-1) = 0
7a+b+5c = 0 ………(2)
Since plane (1) is perpendicular to the plane 2x+6y+6z = 9
a(2) + b(6) + c(6) = 0
2a + 6b +6c = 0
a + 3b + 3c = 0 ………(3)
from eq (2) and eq (3)
= =
= =
= =
Let = = = k
Then a = 3k, b= 4k and C= -5k
From eq (1)
3k(x-2) + 4k(y-2) + (-5k)(z-1) = 0
3x - 6 +4y - 8 - 5z + 5 = 0
3x + 4y - 5z - 9 = 0
This is the required equation of plane.

So, the correct answer is (D): 3x + 4y - 5z - 9 = 0