[tex]\mathbf f(x,y,z)=(x+y^2)\,\mathbf i+(y+z^2)\,\mathbf j+(z+x^2)\,\mathbf k[/tex]
[tex]\nabla\times\mathbf f(x,y,z)=-2z\,\mathbf i-2x\,\mathbf j-2y\,\mathbf k[/tex]
By Stokes' theorem, we have
[tex]\displaystyle\int_C\mathbf f(x,y,z)\cdot\mathrm d\mathbf r=\iint_S\nabla\times\mathbf f(x,y,z)\cdot\mathrm dS[/tex]
where [tex]S[/tex] is the triangle with the given vertices. We can parameterize [tex]S[/tex] by
[tex]\mathbf s(u,v)=(7-7u-7v-7uv)\,\mathbf i+(7v-7uv)\,\mathbf j+7u\,\mathbf k[/tex]
Then the surface integral reduces to
[tex]=\displaystyle-2\int_{u=0}^{u=1}\int_{v=0}^{v=1}(z(u,v)\,\mathbf i+x(u,v)\,\mathbf j+y(u,v)\,\mathbf k)\cdot\left(\frac{\partial\mathbf s}{\partial u}\times\frac{\partial\mathbf s}{\partial v}\right)\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle-2\int_{u=0}^{u=1}\int_{v=0}^{v=1}(686 + 686 u - 1372 u^2 - 2744 u v + 1372 v^2 - 1372 u v^2)\,\mathrm dv\,\mathrm du[/tex]
[tex]=\dfrac{343}3[/tex]
