Answer :

Plane passes through (-1,1,1) and (1,-1,1),

A(x + 1) + B(y-1) + C(z-1)=0 (1)


A(x-1) + B(y + 1) + C(z-1)=0 (2)


Subtracting (1) from (2),


A(x-1-x-1) + B(y + 1-y + 1)=0


-2A + 2B=0


A=B (3)


Now plane is perpendicular to x + 2y + 2z=5


A + 2B + 2C=0 (4)


Using (3) in (4)


B + 2B + 2C=0


3B + 2C=0



Putting values in equation (1)



B(2(x + 1) + 2(y-1)-3(z-1)=0


2x + 2y-3z + 2-2-3=0


2x + 2y-3z-3=0


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