= 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 $\int_C F\cdot dr=\Phi(r_{\rm end})-\Phi(r_{\rm start})$. The endpoints are $(3,0,0)$ and $(0,2,-1)$, so
$$
\boxed{\int_C F\cdot dr=(-12+4)-9=-17.}
$$
The complicated intermediate path does not affect the work. Directly, $\nabla\times F=0$, consistent with the globally defined <potential function>.
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