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).

Explanation

Let the required equation of the plane be
`(3x-y+2z-4)+lambda(x+y+z-2)=0` for some real value of `lambda`
`rArr (3+lambda)x+(-1+lambda)y+(2+lambda)z -(4 +2lambda)=0`…………..(i)
It is given that the point A(2,2,1) lies in (i)
`therefore (3+lambda) xx 2+(-1+lambda) xx 2+(2 +lambda) xx 1-(4+2lambda)=0`
`rArr (6+2lambda)+(-2+2lambda)+(2+lambda)-4-2lambda=0`
`rArr 3lambda=-2 rArr lambda=-2/3`.
Putting, `lambda=-2/3` in (i), we get
`(3-2/3)x+(-1-2/3)y+(2-2/3)z-4+4/3=0`
`rArr (7x)/3-(5y)/3+(4z)/3-8/3=0 rArr 7x-5y+4z-8=0`.
Hence, the required equation of the plane is `7x-5y+4z-8=0`.