Find the equation of the plane passing through the points `(-1,1,1)` and `(1,-1,1)` and perpendicular to the plane `x+2y+2z=5.`

Explanation

The equation of any plane which passes through (-1, 1, 1) is
`" "a(x+1)+b(y-1)+c(z-1)=0" "`(i)
This plane will pass through (1, -1, 1) if
`" "2a-2b=0 or a=b" "` (ii)
Next, (i) will perpendicular to `x+2y+2z=5` if
`" "a:b:c= a:a:((-3)/(2))a=2 : 2: -3`
Putting these values in (i), we get `2(x+1)+2(y-1)-3(z-1) =0 or 2x+2y-3z+3=0`, which is the equation of the required plane.