I'm pretty sure your answer of 9/2 is correct...
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=3v(1-u)\,\vec\imath+3uv\,\vec\jmath+3(1-v)\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex].
In case youre' not sure where this came from: recall the standard parameterization for a line segment connecting two points [tex]\vec p_1[/tex] and [tex]\vec p_2[/tex],
[tex]\vec r(u)=(1-u)\vec p_1+u\vec p_2[/tex]
with [tex]0\le u\le1[/tex]. Then treating [tex]\vec r(u)[/tex] as a "point", we can parameterize a plane containing [tex]\vec r(u)[/tex] and another point [tex]\vec p_3[/tex] by using
[tex]\vec s(u,v)=(1-v)\vec p_3+u\vec r(u)[/tex]
with [tex]0\le v\le 1[/tex].
Now, take the normal vector to [tex]S[/tex] to be
[tex]\vec r_v\times\vec r_u=9v\,\vec\imath+9v\,\vec\jmath+9v\,\vec k[/tex]
Then the flux of [tex]\vec F[/tex] across [tex]S[/tex] is
[tex]\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^1\int_0^19v(3-3v)\,\mathrm du\,\mathrm dv[/tex]
(since the first two components of [tex]\vec F[/tex] cancel in the dot product)
[tex]=\displaystyle27\int_0^1(v-v^2)\,\mathrm dv=\boxed{\frac92}[/tex]
