I'll give my idea here:
A normal to our surface is $\,(1,1,1)\,$ , but we're asked in downward orientation, which I translate as towards the inside of the tetrahedron determined by the plane and the three planes in the first octant, thus we need the normal $n:=(-1,-1,-1)\Longrightarrow\,\, \stackrel \wedge n:=\frac{n}{||n||}=\frac{n}{\sqrt 3} $ and putting $\,D:=\{(x,y)\;:\;0\leq x\leq 1\,,\,0\leq y\leq 1-x\}\,\,,\,dS=\sqrt 3\, d\mathbf A=\sqrt 3\,dxdy\,$ , and since $\,z=1-x-y\,$ , we get
$\int\int_S\mathbf{\vec F}\cdot d\mathbf{\vec S}=\int\int_D\mathbf{\vec F}\cdot \mathbf {\stackrel {\wedge} n}\cdot d\mathbf S=\int_0^1dx\int_0^{1-x}(xze^y\,,\,-xze^y\,,\,z)\cdot (-1,-1,-1)dy=$ $=\int_0^1dx\int_0^{1-x}-(1-x-y)\,dy=\int_0^1\left[-(x-1)^2\left.+\frac{y^2}{2}\right|_0^{1-x}\right]=$ $\int_0^1\left(-(x-1)^2+\frac{(x-1)^2}{2}\right)dx=-\left.\frac{1}{6}(x-1)^3\right|^1_0=-\frac{1}{6}$