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

The equaiton of the plane passing thorugh the intesection of the planes `3x + 2y - z + 1 =0` and `x + y +z - 2 = 0` is
`(3x + 2y -z+1)+ lambda(x+y+z - 2) = 0" ".......(1)`
Now, this plane passes through the point (2,2,1)
`therefore" "[3(2) +2(2) -1+1] + lambda (2+2+1-2) = 0`
`10 + 3lambda = 0`
`therefore" "lambda = (-10)/(3)`
Putting the value of `lambda` in equation (1) , we get
`(3x + 2y -z+1) (10)/(3) (x+y +z -2) = 0`
`rArr" " 9x + 6y - 3z + 3 - 10y - 10z + 20 = 0`
`rArr" "x + 4y + 13z - 23 =0`
Which is the equaiton of required plane.