Normal vector (2x, 2y, 2z),

At point m (1, 2, 1), n 1 = (1, 2,1);

xy-2z = 0

Normal vector (y, x, -2),

At point m (1, 2, 1), N2 = (2, 1, 2);

The tangent direction vector t = n1× N2 = (-5,4,3), that is, (5,4,3).

The tangent equation (x-1)/5 = (y-2)/(-4) = (z-1)/3.

The normal plane equation 5 (x-1)-4 (y-2)+3 (z-1) = 0.

That is 5x-4y+3z =0.