By Stokes' theorem, the line integral of [tex]\vec F[/tex] over [tex]C[/tex] is given by the surface integral of the curl of [tex]\vec F[/tex] over [tex]S[/tex], where [tex]S[/tex] is the region of intersection of the plane [tex]z=y+6[/tex] and the cylinder [tex]x^2+y^2=1[/tex] with [tex]S[/tex] having positive/upward orientation.
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k[/tex]
The curl of [tex]\vec F[/tex] is
[tex]\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\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^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}[/tex]
