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Find the equation of the plane throug...

Find the equation of the plane through the intersection of the planes `3x" "" "y" "+" "2z" "" "4" "=" "0` and `x" "+" "y" "+" "z" "" "2" "=" "0` and the point (2, 2, 1).

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Eqution of a plane throught the intersection of plane `3x-y+2z-4=0` and `x+y+z-2 = 0` is
`(3x-y+2z-4) + lambda (x+y+z-2) = 0 "…….."(1)`
This plane passes through the point `(2,2,1)`.
`:. (3xx22+2xx1) + lambda(2+2+1-2) = 0`
`rArr (6-4)+3lambda=0 rArr 2+3lambda = 0 , lambda = (-2)/(3)`
Put this value of `lambda` in equation ` (1)`
` (3x-y+2z-4) - (2)/(3)(x+y+z-2)=0`
`rArr 9x-3y+6z-12-2x-2y-2z+4=0`
`rArr 7x-5y+4z-8=0`
which is the required equation of plane.
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