Question

Find the equation of the plane through the intersection of the planes and and the point (2, 2, 1)

Answer 1

It is given that the planes ( 3x−y+2z−4=0 ) and ( x+y+z−2=0 ) passes through the point ( 2,2,1 ).

The equation of the plane through the given intersection of planes is,

( 3x−y+2z−4 )+λ( x+y+z−2 )=0(1)

Since the plane passes through the point ( 2,2,1 ), so, it satisfies equation (1).

( 3( 2 )−2+2( 1 )−4 )+λ( 2+2+1−2 )=0 2+3λ=0 3λ=−2 λ=− 2 3

Substitute λ=− 2 3 in equation (1).

( 3x−y+2z−4 )− 2 3 ( x+y+z−2 )=0 9x−3y+6z−12−2x−2y−2z+4 3 =0 7x−5y+4z−8=0

Therefore, the required equation of the plane is 7x−5y+4z−8=0.