Evaluate \int {F.dS} where, F=xi+(z^2-zx)j-xyk and 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 :  P=2x+2y+z-4

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

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

Projecting the triangle plane on xy plane,


\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 \frac{10}{3}

Please let me know where my logic is wrong.

Thank you.