= Solution
Choose the <oriented surface> parametrization $\mathbf r(x,y)=(x,y,1-x)$, with $0\leq x,y\leq1$. Then $\mathbf r_x\times\mathbf r_y=(1,0,1)$ and $\nabla\times\mathbf A=(-3z,x^2,y)$. The <surface integral> is
$$
\iint_S(\nabla\times\mathbf A)\cdot d\mathbf S=\int_0^1\int_0^1[-3(1-x)+y]\,dx\,dy=-1.
$$
For the compatible boundary direction, use successive vertices $(0,0,1)$, $(1,0,0)$, $(1,1,0)$, $(0,1,1)$ and back. The four <line integrals> are respectively
$$
\int_0^1x^2(1-x)\,dx=\frac1{12},\qquad \int_0^1y\,dy=\frac12,\qquad \int_1^0[x^2(1-x)+3(1-x)]\,dx=-\frac{19}{12},\qquad0.
$$
Their sum is \b[$-1$, agreeing with the surface integral]. Choosing the opposite normal and reversing the boundary direction gives $+1$ on both sides. Stating an orientation is necessary because the question does not prescribe one.
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