= Solution
With those constants, integrate $\phi_x=x+3y+3z$ with respect to $x$ to obtain $\phi=x^2/2+3xy+3xz+g(y,z)$. Matching $\phi_y=3x+y-z$ gives $g=y^2/2-yz+h(z)$. Finally, matching $\phi_z=3x-y-2z$ gives $h=-z^2+C$. Thus a <potential of a conservative vector field> is
$$
\boxed{\phi(x,y,z)=\frac{x^2}{2}+3xy+3xz+\frac{y^2}{2}-yz-z^2+C,\qquad \mathbf F=\nabla\phi.}
$$
This explicit polynomial establishes a global potential on all of $\mathbb R^3$; no topological obstruction is present. Its additive constant is arbitrary.
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