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The field is smooth on the simply connected domain ${\mathbb{R}}^3$. Its curl is $\nabla \times \mathbf{F}=(2\lambda -2,2\mu ,0)$, so a necessary and sufficient condition for a conservative vector field isAt those values it is explicitly the gradient of $\Phi =xy-x^2/2-yz+z^2/2$. Equality of the two previous path integrals would impose only $\lambda -\mu =1$, which is weaker: coinciding path integrals need not imply a conservative field. For example $\lambda =2,\mu =1$ makes those two integrals equal but leaves nonzero curl.

Direct differentiation gives $\boxed{\Phi =x^3{y}^{2}z+C}$ with $\nabla \Phi =\mathbf{G}$, so $\mathbf{G}$ is a conservative vector field and its integral depends only on the endpoints. In contrast,is nonzero, for example at $(1,1,1)$. Thus $\mathbf{F}$ is not conservative. Both earlier paths lie in the surface $y=z$, where the two fields happen to coincide. This illustrates why coinciding path integrals need not imply a conservative field.