The part of the plane [tex]z=18-3x-9y[/tex] in the first octant is one of the faces of a tetrahedron whose vertices correspond exactly to the intercepts of the plane. These are
[tex]3x+9\cdot0+0=18\implies x=6\implies(6,0,0)[/tex]
[tex]3\cdot0+9y+0=18\implies y=2\implies(0,2,0)[/tex]
[tex]3\cdot0+9\cdot0+z=18\implies z=18\implies(0,0,18)[/tex]
We can parameterize this surface with the vector-valued function
[tex]\mathbf s(u,v)=(6(1-u)(1-v),2u(1-v),18v)[/tex]
where [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then the surface element is
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=12\sqrt{91}(1-v)\,\mathrm du\,\mathrm dv[/tex]
so the surface integral becomes
[tex]\displaystyle\iint_{\mathcal S}e^z\,\mathrm dS=12\sqrt{91}\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-v)e^{18v}\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle12\sqrt{91}\int_{v=0}^{v=1}(1-v)e^{18v}\,\mathrm dv[/tex]
[tex]=\dfrac{12\sqrt{91}}{324}e^{18v}(19-18v)\bigg|_{v=0}^{v=1}[/tex]
[tex]=\dfrac{\sqrt{91}}{27}(e^{18}-228)[/tex]
