Stokes' theorem equates the line integral of [tex]\vec F[/tex] along the curve to the surface integral of the curl of [tex]\vec F[/tex] over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).
Parameterize this triangle (call it [tex]T[/tex]) by
[tex]\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)[/tex]
[tex]\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]T[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)[/tex]
Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of [tex]T[/tex] is traversed in the counterclockwise direction when viewed from above.
Compute the curl of [tex]\vec F[/tex]:
[tex]\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)[/tex]
Then by Stokes' theorem,
[tex]\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S[/tex]
where
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS[/tex]
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
The integral thus reduces to
[tex]\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}[/tex]
