= Solution
Write the plane as $z=6-3x-2y$. Its projection on the $xy$ plane is the triangle $0\leq x\leq2$, $0\leq y\leq3-3x/2$. The scalar <surface area element> is
$$
d\Sigma=\sqrt{1+z_x^2+z_y^2}\,dx\,dy=\sqrt{14}\,dx\,dy.
$$
Thus the weighted <surface integral> is
$$
\boxed{\iint_\Sigma y\,d\Sigma=\sqrt{14}\int_0^2\int_0^{3-3x/2}y\,dy\,dx=3\sqrt{14}.}
$$
A geometric check uses the triangular face's area $3\sqrt{14}$ and mean $y$ coordinate $(0+3+0)/3=1$. Multiplying them gives the same result, because $y$ is an affine function on this planar triangle. Its edges have zero surface measure, so strict or non-strict octant inequalities do not change the integral.
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