Check that the point (-1, -1, 1) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y, z), find a vector normal to the surface and an equation for the tangent plane to the surface at (-1, -1, 1). 2x² - 2y² + 3z² = 2. Vector normal = ?, Tangent plane: ? = z + _

A) Vector normal = (0, 0, 3), Tangent plane: z = -1
B) Vector normal = (2, -2, 3), Tangent plane: z = 1
C) Vector normal = (-2, 2, 3), Tangent plane: z = 0
D) Vector normal = (1, 1, 1), Tangent plane: z = 2