= Solution
On $OAC$, $y=0$ and the outward <unit normal> is $-\mathbf e_y$. Its <flux> is
$$
\begin{aligned}
\Phi_{OAC}&=-\int_0^1\int_0^{1-x}(x+z)^2\,dz\,dx\\
&=-\frac13\int_0^1(1-x^3)\,dx=\boxed{-\frac14}.
\end{aligned}
$$
On $OBC$, $x=0$ and the outward <unit normal> is $-\mathbf e_x$. Hence
$$
\Phi_{OBC}=-\int_0^1\int_0^{1-y}(y+z)\,dz\,dy
=-\frac12\int_0^1(1-y^2)\,dy=\boxed{-\frac13}.
$$
On $ABC$, write $z=1-x-y$ over $x\geq0$, $y\geq0$, $x+y\leq1$. The <oriented surface element> is $(1,1,1)^Tdx\,dy$, pointing away from the origin. Here $x+y+z=1$, so $\mathbf F=(1,1,0)^T$. Thus
$$
\Phi_{ABC}=\int_0^1\int_0^{1-x}2\,dy\,dx=\boxed{1}.
$$
The unnormalized <normal vector> in the <oriented surface element> already incorporates the factor $\sqrt3$ from the surface <area>; no additional factor is needed.
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