= Solution
The plane cuts the positive coordinate axes at $(3,0,0)$, $(0,3,0)$ and $(0,0,3/2)$. Choose its <unit normal vector> $(1,1,2)/\sqrt6$, pointing away from the origin. In the parametrization $(x,3-x-2z,z)$, the <vector surface element> for this orientation is $(1,1,2)\,dx\,dz$. Thus
$$
\mathbf A\cdot d\mathbf S=(x,-x,z)\cdot(1,1,2)\,dx\,dz=2z\,dx\,dz,
$$
and the triangular projection gives
$$
\boxed{\iint_S\mathbf A\cdot d\mathbf S=\int_0^3\int_0^{(3-x)/2}2z\,dz\,dx=\frac94.}
$$
For the opposite orientation the answer is $-9/4$. The normal direction must be stated because this open surface has no prescribed orientation in the question.
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