**Given Information:**

The vector field F is given as:

F = (2x^2 - 3z) î - 2xy ĵ - 4x k̂

The region V is bounded by the planes x = 0, y = 0, z = 0, and 2x + 2y + z = 4.

**Evaluation of Triple Integral of ∇F:**

To evaluate the triple integral of ∇F, we need to find ∇F first.

The gradient of a vector field F is given by:

∇F = (∂F/∂x) î + (∂F/∂y) ĵ + (∂F/∂z) k̂

Taking partial derivatives of F with respect to x, y, and z:

∂F/∂x = 4x î - 2y ĵ - 4 k̂
∂F/∂y = -2x ĵ
∂F/∂z = -3 î

Therefore, ∇F = (4x î - 2y ĵ - 4 k̂) î - (2x ĵ) ĵ - 3 î = (4x - 3) î - 2xy ĵ - 4x k̂

**Defining the Region V:**

The region V is bounded by the planes x = 0, y = 0, z = 0, and 2x + 2y + z = 4.

To visualize the region V, we can look at the intersection of these planes. The plane x = 0 represents the yz-plane, the plane y = 0 represents the xz-plane, and the plane z = 0 represents the xy-plane. The equation 2x + 2y + z = 4 can be rewritten as z = 4 - 2x - 2y.

Considering these equations, we can see that the region V is a tetrahedron with vertices at (0, 0, 0), (2, 0, 0), (0, 2, 0), and (0, 0, 4).

**Evaluating the Triple Integral:**

The triple integral of ∇F over the region V can be written as:

∫∫∫ (∇F) dV

Using the divergence theorem, this can be rewritten as:

∫∫∫ ∇⋅F dV

The divergence of F is given by:

∇⋅F = ∂(4x - 3)/∂x + ∂(-2xy)/∂y + ∂(-4x)/∂z
= 4 - 2y

Now, we need to evaluate the triple integral of (4 - 2y) over the region V.

Using the given bounds for x, y, and z, the triple integral becomes:

∫(0 to 4) ∫(0 to 2 - x/2) ∫(0 to 4 - 2x - 2y) (4 - 2y) dz dy dx

Solving the