The surface (call it [tex]S[/tex]) is a triangle with vertices at the points
[tex]x=0,y=0\implies z=2\implies(0,0,2)[/tex]
[tex]x=0,y=2\implies z=-2\implies(0,2,-2)[/tex]
[tex]x=2,y=0\implies z=-8\implies(2,0,-8)[/tex]
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_v\times\vec s_u=(20v,8v,4v)[/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 s_v\times\vec s_u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}[/tex]
