Consider the function F (x, y, z) = 2x5 + 3x2y − 2xz + y6 + z2 − 6.(a) Find a vector point-normal and scalar equation and a vector parametric form for thetangent plane to the surface F (x, y, z) = 0 at the point p = (−1, 0, 2).

To find the tangent plane to the surface F(x, y, z) = 0 at the point p = (-1, 0, 2), we first need to find the gradient of F at p, which will give us the normal vector to the plane.

The gradient of F is given by the vector of its partial derivatives:

∇F = (F_x, F_y, F_z) = (10x^4 + 6xy - 2z, 6y^5 + 3x^2, 2z - 2x)

Evaluating ∇F at p = (-1, 0, 2), we get:

∇F(p) = (10(-1)^4 + 6(-1)(0) - 2(2), 6(0)^5 + 3(-1)^2, 2(2) - 2(-1)) = (8, 3, 6)

So, the normal vector to the plane is (8, 3, 6).

The point-normal form of the equation of the plane is given by:

n . (r - r_0) = 0

where n is the normal vector, r = (x, y, z) is a point on the plane, and r_0 = p is a known point on the plane. Substituting the known values, we get:

(8, 3, 6) . ((x, y, z) - (-1, 0, 2)) = 0

which simplifies to:

8(x + 1) + 3y + 6(z - 2) = 0

This is the scalar equation of the plane.

The vector parametric form of the equation of the plane can be obtained by expressing (x, y, z) in terms of two parameters, say s and t, as follows:

(x, y, z) = (-1, 0, 2) + s(1, 0, 0) + t(0, 1, 0)

This is the vector parametric form of the equation of the plane.