 Solution

The field is the gradient of $\Phi =x^2-3y^2+4z^2$ on all of ${\mathbb{R}}^3$. It is therefore a conservative vector field, and the fundamental theorem for line integrals gives ${\int }_{C}F\cdot dr=\Phi (r_{\mathrm{end}})-\Phi (r_{\mathrm{start}})$. The endpoints are $(3,0,0)$ and $(0,2,-1)$, soThe complicated intermediate path does not affect the work. Directly, $\nabla \times F=0$, consistent with the globally defined potential function.