[tex]\displaystyle\int_C(z+y)\,\mathrm dx+(2x-z)\,\mathrm dy+(x-y)\,\mathrm dz[/tex]
has the underlying vector field
[tex]\vec F(x,y,z)=(z+y)\,\vec\imath+(2x-z)\,\vec\jmath+(x-y)\,\vec k[/tex]
with curl
[tex]\nabla\times\vec F(x,y,z)=\vec k[/tex]
Parameterize [tex]T[/tex] by
[tex]\vec s(u,v)=(1-v)((1-u)\,\vec\imath+u\,\vec\jmath)+v(2\,\vec k)[/tex]
[tex]\vec s(u,v)=(1-v)(1-u)\,\vec\imath+u(1-v)\,\vec\jmath+2v\,\vec k[/tex]
with [tex]0\le u\le1[/tex], [tex]0\le v\le1[/tex].
The implied orientation of the curve is counter-clockwise when viewing [tex]T[/tex] from the first octant (each of [tex]x,y,z[/tex] greater than 0), so that the normal vector to [tex]T[/tex] is pointing upward and away from the origin. Take this vector to be
[tex]\vec n=\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2(1-v)\,\vec\imath+2(1-v)\,\vec\jmath+(1-v)\,\vec k[/tex]
Then by Stokes' theorem, the line integral is equivalent to
[tex]\displaystyle\iint_T(\nabla\times\vec F(x,y,z))\cdot\mathrm d\vec S=\iint_T\vec k\cdot\vec n\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^1\int_0^1(1-v)\,\mathrm du\,\mathrm dv=\boxed{\frac12}[/tex]
