Solution
= Solution
For the second <vector field>, the <curl> is
$$
\boxed{\nabla\times\mathbf G=\bigl(2y-2z,\ 2z-2x,\ 2x-2y\bigr).}
$$
It is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable <scalar potential> would have zero <curl> of its <gradient>, so \b[$\mathbf G$ is not conservative]. The fact that its <curl> vanishes on the line $x=y=z$ does not make it vanish throughout the domain.