= 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 $x=1-t$, $y=t$, $z=0$, $0\le t\le1$. Then
$$
A\cdot\frac{d\mathbf x}{dt}=(-t^2,(1-t)^2,t^2-(1-t)^2)\cdot(-1,1,0)=t^2+(1-t)^2.
$$
Its integral is $2/3$, so the required <circulation> is
$$
\boxed{\oint_C A\cdot d\mathbf x=3\cdot\frac23=2}.
$$
For a sufficiently smooth <vector field> on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the <Stokes theorem> states
$$
\int_T(\nabla\times A)\cdot n\,dS=\oint_C A\cdot d\mathbf x.
$$
Therefore the requested <surface integral>, with all normal components positive, is also \b[$2$].
To verify it directly, parametrize the triangle by $\mathbf r(y,z)=(1-y-z,y,z)$ on $y\ge0$, $z\ge0$, $y+z\le1$. Its two tangent vectors are $\mathbf r_y=(-1,1,0)$ and $\mathbf r_z=(-1,0,1)$, giving the <vector area element>
$$
\boxed{d\mathbf S=(\mathbf r_y\times\mathbf r_z)\,dy\,dz=(1,1,1)\,dy\,dz}.
$$
Differentiation 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 $4$. Consequently
$$
\int_T(\nabla\times A)\cdot d\mathbf S=\int_0^1\int_0^{1-y}4\,dz\,dy=2,
$$
confirming both the magnitude and the orientation sign.
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