Q. 154.0( 1 Vote )

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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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