Let C be the triangle with vertices $(5, 0, 0)$, $(0, 5, 0)$ and $(0, 0, 2)$. I want to find the line integral $$\oint_C \frac{4}{25}xydx+\frac{2}{5}yzdy+1xzdz \,$$ where $C$ is the curve oriented clockwise as seen from the point $(5, 5, 5)$.
I found the projection of my triangle in the $xy$-plane to be limited by $$0\leq x \leq 5$$ and $$0 \leq y \leq 5-x$$
I also found $curl F$ = $[-\frac{2}{5}y, -z, -\frac{4}{25}x]$. The plane intersecting the three points is $2x+2y+5z-10=f(x,y,z)$ and thus $\nabla f = [2, 2, 5]$
Substituting $-z$ where $z=-\frac{2}{5}(x+y-5)$ and taking the dot product $curl F \cdot \nabla f$ I am left with $-4$ and $$ \int_{0}^{5}\int_{0}^{5-x} -4 dy dx = -50$$ but this is not the correct answer. What have I done wrong?