Answer :
Plane passes through (1,1,2) and (2,-2,2),
A(x-1) + B(y-1) + C(z-2)=0 (1)
A(x-2) + B(y + 2) + C(z-2)=0 (2)
Subtracting (1) from (2),
A(x-2-x + 1) + B(y + 2-y-1)=0
A-3B=0 (3)
Now plane is perpendicular to 6x-2y + 2z=9
6A-2B + 2C=0 (4)
Using (3) in (4)
18A-2B + 2C=0
16B + 2C=0
C=-8B
Putting values in equation (1)
3B(x-1) + B(y + 2)-8B(z-2)=0
B(3x-3 + y + 2-8z + 16)=0
3x + y-8z + 15=0
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