(d) Find the equations of the tangent plane and normal line to the surface 4 = xyz at the point i+2j+2k. -----------------------------------------

Hello student,
The equation of surface is F(x, y, z) = xyz – 4 = 0
Differentiating this equation partially w.r.t. x, y and z respectively
∂F/∂x = yz
∂F/∂y = xz
∂F/∂z = xy
Hence, at the point (1, 2, 2) we have
∂F/∂x = 4
∂F/∂y = 2
∂F/∂z = 2
Hence, the equation of tangent plane at (1, 2, 2) is
(x-1)4 + (y-2)2 + (z-2)2 = 0
So, 2x + y+z = 6
Equation of normal line at point (1, 2, 2) is
(x-1)/4 = (y-2)/2 = (z-2)/2
or (x-1)/2= (y-2)/1 = (z-2)/1