A

In classical mechanics, the force acting on a particle is related to the potential energy by the equation:
\[ \vec{F} = -\nabla U \]
where \( \nabla U \) is the gradient of the potential energy function.
Step 1: Calculate the partial derivatives of U with respect to x, y, and z:
\[ U = 2x^2 + 3y^3 + 2z \]
\[ \frac{\partial U}{\partial x} = 4x \]
\[ \frac{\partial U}{\partial y} = 9y^2 \]
\[ \frac{\partial U}{\partial z} = 2 \]
Step 2: Evaluate the partial derivatives at the point P (1, 2, 3):
\[ \frac{\partial U}{\partial x}\bigg|_{(1,2,3)} = 4(1) = 4 \]
\[ \frac{\partial U}{\partial y}\bigg|_{(1,2,3)} = 9(2^2) = 36 \]
\[ \frac{\partial U}{\partial z}\bigg|_{(1,2,3)} = 2 \]
Step 3: Calculate the force components:
\[ F_x = -\frac{\partial U}{\partial x} = -4 \]
\[ F_y = -\frac{\partial U}{\partial y} = -36 \]
\[ F_z = -\frac{\partial U}{\partial z} = -2 \]
Step 4: Thus, the force acting on the particle at point P (1m, 2m, 3m) is:
\[ \vec{F} = (-4, -36, -2) \, \text{N} \]
Therefore, the correct answer is the force vector: \( \vec{F} = (-4, -36, -2) \, \text{N} \).