Q. 175.0( 2 Votes )

Find the equation

Answer :

Plane passes through (3,4,2) and (7,0,6),

A(x-3) + B(y-4) + C(z-2)=0 (1)


A(x-7) + B(y-0) + C(z-6)=0 (2)


Subtracting (1) from (2),


A(x-7-x + 3) + B(y-y + 4) + C(z-6-z + 2)=0


-4A + 4B-4C=0


A-B + C=0


B=A + C (3)


Now plane is perpendicular to 2x-5y=15


2A-5B=0 (4)


Using (3) in (4)


2A-5(A + C)=0


2A-5A-5C=0


-3A-5C=0




Putting values in equation (1)



A(5(x-3) + 2(y-4)-3(z-2)=0


5x + 2y-3z-15-8 + 6=0


5x + 2y-3z-17=0


So, required equation of plane is 5x + 2y-3z-17=0.


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