Parameterize [tex]S[/tex] by [tex]\vec s(u,v)=(u,v,1+2u+3v)[/tex] with [tex]0\le u\le4[/tex] and [tex]0\le v\le2[/tex]. Then the surface element is
[tex]\mathrm dS=\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\sqrt{14}\,\mathrm du\,\mathrm dv[/tex]
so that the integral is
[tex]\displaystyle\iint_Sx^2yz\,\mathrm dS=\sqrt{14}\int_0^2\int_0^4u^2v(1+2u+3v)\,\mathrm du\,\mathrm dv=\boxed{\frac{1408\sqrt{14}}3}[/tex]
