= Solution
For an oriented piecewise smooth <surface> $S$, a continuously differentiable <vector field> $F$ on a neighborhood of it, and its compatibly oriented boundary $C=\partial S$, <Stokes theorem> states
$$
\boxed{\oint_CF\cdot d r=\iint_S(\nabla\times F)\cdot n\,dS.}
$$
Here $n$ is the chosen unit <normal vector>, $dS$ is the unsigned <surface area element>, and $d r$ is the tangent displacement along $C$. The <right-hand rule> relates the boundary direction to $n$.
The supplied field has <curl> $\nabla\times F=(2z,2x,2y)$. Parametrize the plane by $r(x,y)=(x,y,1-x-y)$ over the triangle $D=\{x,y\geq0:x+y\leq1\}$. Its upward oriented <surface area element> is $(1,1,1)\,dx\,dy$, giving
$$
\iint_S(\nabla\times F)\cdot d\boldsymbol S
=\int_D2(x+y+z)\,dx\,dy
=2\operatorname{area}(D)=\boxed{1}.
$$
For the boundary <line integral>, take the positively oriented vertices $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Their projections run counterclockwise in the horizontal plane. With $0\leq t\leq1$, the three edges are
$$
r_{AB}=(1-t,t,0),\qquad r_{BC}=(0,1-t,t),\qquad r_{CA}=(t,0,1-t).
$$
On each edge, direct substitution gives $F(r(t))\cdot r'(t)=t^2+2t-1$. Hence
$$
\boxed{\oint_CF\cdot d r
=3\int_0^1(t^2+2t-1)\,dt=3\left(\frac13\right)=1.}
$$
The two independent computations agree, including the orientation sign.
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