Through the point (2,1,2), and respectively perpendicular to the plane x + 3Y + Z = 2, and the plane 3x + 2y-4y = 1, the plane equation is solved

Through the point (2,1,2), and respectively perpendicular to the plane x + 3Y + Z = 2, and the plane 3x + 2y-4y = 1, the plane equation is solved

From x + 3Y + Z = 2,3x + 2y-4z = 1
x+3y=2-z,3x+2y=1+4z,
The solution is (x + 1 / 7) / 2 = (Y-5 / 7) / (- 1) = Z, which is a two-sided intersection equation,
The plane is perpendicular to two known surfaces, so it is perpendicular to their intersection,
The normal vector of the plane is parallel to the direction vector of the intersection line,
The plane equation is 2 (X-2) - (Y-1) + (Z-2) = 0,
That is, 2x-y + Z-5 = 0