Simplification on stokes theorem problem we have a problem , but is more conceptual , than getting the answer, is from larson calculus 9, 15.8.11 and says:
$\ F(x,y,z) = 2y\mathbf{i} + 3z\mathbf{j} + x\mathbf{k}$
$\ C: triangle\ with\  vertex\ (2,0,0)\ (0,2,0)\ (0,0,2)$
the answer is not the problem, cause it is -12, the problem is here
given that:
$\ \int_c \mathbf{F}\cdot d\mathbf{r} = \int_s\int (curl\ \mathbf{F})\cdot \mathbf{N}\ dS $
so
$\ curl\ F =\ <-3,-1,-2> $
given the points, we can find two vector, in order to find  $\ \mathbf{N} $
$\ \vec{A} = (2,0,0)-(0,2,0) $
$\ \vec{A} =\ <2,-2, 0 > $
$\ \vec{B} = (2,0,0)-(0,0,2) $
$\ \vec{B} =\ <2,0, -2 > $
$\ \textbf{N} = \vec{A} \times \vec{B}  = <4,4,4>$
how ever if we divide it by 4, we also have
$\   \textbf{N}_{1}  = <1,1,1>$
since $\ \textbf{N},  \textbf{N}_{1} $ has to be an unit vector we divide by his magnitude
$\  || \textbf{N} || \ = 4 \sqrt 3  $ 
$\ \frac{\textbf{N}}{|| \textbf{N} ||} = \  \frac{1}{4\sqrt 3} \cdot <4,4,4> $
$\  || \textbf{N}_{1} || \ = \sqrt 3  $ 
$\ \frac{\textbf{N}_{1}}{|| \textbf{N}_{1} ||} = \  \frac{1}{\sqrt 3} \cdot <1,1,1> $
here is the problem
$\ dS = || r_{u} \times r_{v}  || dA $
$\  \textbf{N} =  r_{u} \times r_{v}  $
$\  dS = || \textbf{N} ||dA  $
$\ \int_{D}\int <-3,-1,-2> \cdot\  \frac{1}{4 \sqrt 3} \cdot <4,4,4> \cdot\ 4 \sqrt 3\   dA  = -24 $
$\ \int_{D}\int <-3,-1,-2> \cdot\  \frac{1}{\sqrt 3} \cdot <1,1,1> \cdot\  \sqrt 3\  dA = -12 $ 
I dont know why the answer is different if mathematically both are the same
D = projection in the xy plane, but is not necessary because is the shape that is reflexed on the xy plane is a triangle
 A: If you use $\ dS = \Vert r_{u} \times r_{v}\Vert dA$ you have to tell what is the surface parametrization using $u$ and $v$. As you drive the last integrals (integrating in the projection of the surface in the plane $xy$), the parametrization must be $u=x$ and $v=y$. Anyway, you didn't calculate the cross product; you simply used two versions of the normal unit vector and some factors. Why the factors aren't the same? at last, the unit vectors you are using are the same!
Calculating first the unit vector has exactly that problem, that you miss what is the companion factor for $dA$. If you insist into calculating this way and with the parametrisation with $x$ and $y$, the factor is the ratio between the area of the tilted triangle and that of its projection on the $xy$ plane. The cross product way is the better one.
The surface is $x+y+z=2$ or $z=2-x-y$
$r(u,v)=\begin{cases}
 x=u\\
 y=v\\
 z=2-u-v
\end{cases}$
$r_u=(1,0,-1)$ and $r_v=(0,1,-1)$
$r_u\times r_v=(1,1,1)$ Then $dS=\sqrt{3}dA$
