The plane 4x + 7y + 4z + 81=0 is rotated through a right angle about its line of intersection with the plane 5x + 3y + 10z =25.The equation of the plane in its new position is

The equation of the plane through the line of intersection of the planes 4x + 7y + 4z + 81 = 0 and 5x + 3y + 10z = 25 is (4x+7y+4z+81)+λ(5x+3y+10z−25)=0or (4+5λ)x+(7+3λ)y+(4+10λ)z+81-25λ=0-----iwhich is perpendicular to 4x + 7y + 4z +81 = 0 ⇒    4(4+5λ)+7(7+3λ)+4(4+10λ)=0 or     81λ+81=0 or      λ=−1Hence, the plane is x - 4y + 6z = 106