Without carrying out any integration, show that the line integral $$\int_{C} \nabla\phi\cdot \mathbf{dr} = 3$$ where $C$ is any smooth curve joining $(-1, 3, 9)$ to $(1, 6,-4)$ and $\phi = xyz$.
I don't really have any idea where to start here, since $C$ is an arbitrary path I can't parametrise it in any way. I know $\nabla\phi = yz\mathbf{i} + xz\mathbf{j}+xy\mathbf{j}$ now Im not really sure where to go.
$$\int_C \nabla\phi \cdot {\rm d}{\bf r} = \phi(1,6,-4)-\phi(-1,3,9) = -24-(-27) = 3.$$