Find the equation of the plane that contains the point A(1,-1,2) and is perpendicular to both the planes `2x+3y-2z=5` and `x+2y-3z=8`. Hence, find the distance of the point `P(-2,5,5)` from the plane obtained above

Answer

`5x-4y-z-7=0, sqrt(42)` units

Explanation

Any plane through A(1,-1,2) is given by
`a(x-1)+b(y-1)+c(z-2)=0`……………..(i)
Since it is perpendicular to each of the planes `2x+3y-2z=5` and `x+2y-3z=8`, we have
`2a+3b-2c=0`………………….(ii)
`a+2b-3c=0`……………….(iii)
On solving (ii) and (iii) by cross multiplication, we have
`a/(-9+4)=b/(-2+6)=c/(4-3)=lambda=a=-5lambda,b=4lambda,c=lambda`.
Putting these values in (i), we get the required equation as
`-5lambda(x-1)+4lambda(y+1)+lambda(z-2)=0`.
`rArr 5(x-1)+4lambda(y+1)-(z-2)=0 rArr 5x-4y-z-7=0`.
Distance of the point P(-2,5,5) from this plane is given by
`d=(|5 xx (-2)-4 xx 5 -5 -7|)/sqrt(5^(2)+(-4)^(2)+(-1)^(2))=(|-42|)/sqrt(42)=sqrt(42)` units.