Find the equation of the plane that contains the point A(1, -1, 2) and is perpendicular to both the planes 3x + 3y – 2z = 5 and x + 2y - 3z = 8. Hence, find the distance of the point P(-2, 5, 5) from the plane obtained above.
Applying condition of perpendicularity between planes,
Where A, B, C are direction ratios of plane and A1, B1, C1 are of other
plane.
(1)
(2)
And plane contains the point (1,-1,2),
A(x-1) + B(y + 1) + C(z-2)=0 (3)
On solving equation (1) and (2)
Putting values in equation (3)
-5x + 5 + 7y + 7 + 3z-6=0
-5x + 7y + 3z + 6=0
5x-7y-3z-6=0
For equation of plane Ax + By + Cz=D and point (x1,y1,z1), distance of a
point from a plane can be calculated as
1