= Solution
Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-\mathbf e_x$ or $-\mathbf e_y$, so their flux of the <curl> is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)\,dx\,dy$, with $x,y\geq0$, $x+y\leq1$. Thus the direct <surface integral> is
$$
\int_S(\nabla\times\mathbf A)\cdot\mathbf n\,dS=\int_0^1\int_0^{1-x}1\,dy\,dx=\frac12.
$$
This surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf A=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral $\int_0^12t\,dt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving $\int_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. Therefore
$$
\boxed{\oint_C\mathbf A\cdot d\mathbf r=1-\frac12+0=\frac12=\int_S(\nabla\times\mathbf A)\cdot\mathbf n\,dS.}
$$
Both sides have been calculated directly, verifying the <Stokes theorem> with consistent orientations.
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