 Solution

Orient the triangular boundary from $P=(1,0,0)$ to $Q=(0,1,0)$ to $R=(0,0,1)$ and back to $P$. This orientation corresponds to the positive normal vector $(Q-P)\times (R-P)=(1,1,1)$. The cyclic coordinate rotation $(x,y,z)\mapsto (z,x,y)$ sends the three directed edges to one another and transforms the vector field by the same rotation. Since rotations preserve dot products, the three edge contributions to the line integral are equal.

On the first edge use $x=1-t$, $y=t$, $z=0$, $0\le t\le 1$. ThenIts integral is $2/3$, so the required circulation is

For a sufficiently smooth vector field on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the Stokes theorem statesTherefore the requested surface integral, with all normal components positive, is also $2$.

To verify it directly, parametrize the triangle by $\mathbf{r}(y,z)=(1-y-z,y,z)$ on $y\ge 0$, $z\ge 0$, $y+z\le 1$. Its two tangent vectors are ${\mathbf{r}}_y=(-1,1,0)$ and ${\mathbf{r}}_z=(-1,0,1)$, giving the vector area elementDifferentiation of the vector field gives $\nabla \times A=2(y+z,z+x,x+y)$. On the triangle $x+y+z=1$, so its dot product with $(1,1,1)$ is $4$. Consequentlyconfirming both the magnitude and the orientation sign.