By Stokes' theorem, the line integral of [tex]\vec F[/tex] over [tex]C[/tex] is equivalent to the surface integral of the curl of [tex]\vec F[/tex] over [tex]S[/tex], where [tex]S[/tex] is the part of the plane [tex]7x+6y+z=1[/tex] in the first octant, with [tex]S[/tex] having positive/upward orientation.
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=\dfrac{(1-u)(1-v)}7\,\vec\imath+\dfrac{u(1-v)}6\,\vec\jmath+v\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_u\times\vec s_v=\dfrac{1-v}6\,\vec\imath+\dfrac{1-v}7\,\vec\jmath+\dfrac{1-v}{42}\,\vec k[/tex]
The curl of [tex]\vec F[/tex] is
[tex]\nabla\times\vec F(x,y,z)=(x-y)\,\vec\imath-y\,\vec\jmath+\vec k[/tex]
Then the line integral is equivalent to
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\int_0^1\int_0^1\left(\frac{(6-13u)(1-v)}{42}\,\vec\imath-\dfrac{u(1-v)}6\,\vec\jmath+\vec k\right)\cdot\left(\dfrac{1-v}6\,\vec\imath+\dfrac{1-v}7\,\vec\jmath+\dfrac{1-v}{42}\,\vec k\right)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\frac1{252}\int_0^1\int_0^1(12-6v-19u+19uv)(1-v)\,\mathrm du\,\mathrm dv=\boxed{\frac{11}{1512}}[/tex]
