Evaluate $\displaystyle \int {F.dS}$ where, $\displaystyle F=xi+(z^2-zx)j-xyk$ and $\displaystyle S$ is a triangular surface with vertices (2,0,0) , (0,2,0) and (0,0,4)

My attempt :

The equation of the triangular plane : $\displaystyle P=2x+2y+z-4$

The normal surface vector : $\displaystyle \nabla {P}=2i+2j+k$

Unit normal surface vector : $\displaystyle \frac{2i+2j+k}{3}$

Projecting the triangle plane on xy plane,

$\displaystyle dS=\frac{dxdy}{\frac{1}{3}}$

$\displaystyle \therefore S=\int_0^2{\int_0^{2-y}(2i+2j+k).(xi+(z^2-zx)j-xyk)dxdy$

As the projection is on xy plane, z=0

On integration, S=2

The correct solution however is $\displaystyle \frac{10}{3}$

Please let me know where my logic is wrong.

Thank you.