A

Given potential energy function:
U(x, y, z) = 2x^2 + 3y^3 + 2z

To find the force, we use the relation:
F = -\nabla U = -\left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right)

Step 1: Calculate the partial derivatives of U.
\( \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x \)
\( \frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(2x^2 + 3y^3 + 2z) = 9y^2 \)
\( \frac{\partial U}{\partial z} = \frac{\partial}{\partial z}(2x^2 + 3y^3 + 2z) = 2 \)

Step 2: Evaluate the derivatives at point P (1m, 2m, 3m).
\( \frac{\partial U}{\partial x}|_{(1,2,3)} = 4(1) = 4 \)
\( \frac{\partial U}{\partial y}|_{(1,2,3)} = 9(2)^2 = 9(4) = 36 \)
\( \frac{\partial U}{\partial z}|_{(1,2,3)} = 2 \)

Step 3: Calculate the force components.
F = -\left( 4, 36, 2 \right) = -4\hat{i} - 36\hat{j} - 2\hat{k}

Therefore, the force acting on the particle at point P (1m, 2m, 3m) is:
F = -4\hat{i} - 36\hat{j} - 2\hat{k}.