Parameterize [tex]C[/tex] by
[tex]\vec r(t)=(1-t)(0,0,0)+t(1,3,6)=(t,3t,6t)[/tex]
with [tex]0\le t\le1[/tex]. Then the line integral is
[tex]\displaystyle\int_Cxyz\,\mathrm dS=\int_0^1x(t)y(t)z(t)\left\|\frac{\mathrm d\vec r}{\mathrm dt}\right\|\,\mathrm dt[/tex]
[tex]=\displaystyle18\sqrt{91}\int_0^1t^3\,\mathrm dt=\boxed{\frac{9\sqrt{91}}2}[/tex]
