The equation of the plane passing through the point (1,2,1) and perpendicular to the planes x+2 y +2 z -7=0 and 3 x+3 y +2 z -5=0 is

Step-by-step solution

Determining the normal vector The normal vector of the plane must be perpendicular to the normals of both given planes. For x+2y+2z-7=0 , the normal is n ₁ = 1, 2, 2 ; for 3x+3y+2z-5=0 , it is n ₂ = 3, 3, 2 . Computing the cross product The required normal vector is N = n ₁ n ₂ = vmatrix i & j & k 1 & 2 & 2 3 & 3 & 2 vmatrix = i (4-6) - j (2-6) + k (3-6) = -2 i + 4 j - 3 k , so A = -2 , B = 4 , C = -3 . Constructing the plane equation Using the point-normal form with point (1,2,1) : -2(x-1) + 4(y-2) - 3(z-1) = 0 ,