= Solution
For $\mathbf F$, all three <curl> components vanish for every $a$. More constructively,
$$
\mathbf F=\nabla\phi_F,\qquad
\phi_F=2x^2+y^2+z^2-2xy-2xz+ayz.
$$
Hence \b[$\mathbf F$ is conservative for every real $a$]. Its endpoint <vector-field potential> difference is $a$, checking both <line integrals>.
For the other <vector field>, direct <partial differentiation> gives
$$
\nabla\times\mathbf G=2b(1,1,1)^T.
$$
It therefore has a <vector-field potential> exactly when $b=0$, with no restriction on $c$:
$$
\boxed{b=0,\quad c\text{ arbitrary},\qquad
\phi_G=\sin(xy)+cxz+C}.
$$
The value of $a$ is irrelevant to $\mathbf G$. Its rotational part for general $b$ is $b(1,1,1)^T\times\mathbf r$. Both selected paths lie in $x=y$, and this rotational part has zero <dot product> with every <tangent vector> to that <plane>. That explains why these paths cannot detect its nonzero <curl> when $b\ne0$.
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