Parametrizing a line integral vector field evaluate the line integral where
$$F = \langle y^2, x, z^2 \rangle$$
and $C$ is the curve of intersection of the plane $x+y+z=1$ and the cylinder $x^2+y^2=1$. orientated clockwise when viewed from above.
I'm a bit confused because I've parametrized the vector field: 
$$F = \langle sin^2(t), cos(t), 1-cos(t)-sin(t) \rangle$$
But when I parametrize $r(t)$:
$$r(t) = \langle cos^2(t), sin^2(t) \rangle$$ 
What about $z$? If I take the dot product of $F$ and $dr$ I will get $0$ in the $z$ component.

You don't parametrize the vector field, you parametrize the curve and then plug in the parametrization of the curve into the vector field. Namely, 
$$ r(t) = (\cos(t), -\sin(t), 1 - \cos(t) + \sin(t)), \,\,\, t \in [0,2\pi] $$
and then
$$ \int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{2\pi} F(r(t)) \cdot r'(t) \, dt = \\
\int_0^{2\pi} (\sin(t)^2, \cos(t), (1 - \cos(t) + \sin(t))^2) \cdot (-\sin(t), -\cos(t), \sin(t) + \cos(t)) \, dt = \\
\int_0^{2\pi} -\sin^3(t) - \cos^2(t) + ((1-\cos(t))^2 + 2(1-\cos(t))\sin(t) + \sin^2(t)) \cdot (\sin(t) + \cos(t)) \, dt =\\
\int_0^{2\pi} -\cos^2(t) - \sin^3(t) + (4 - 4\cos(t))(\sin(t) + \cos(t)) \, dt = \\
\int_0^{2\pi} 5\cos^2(t) - \sin^3(t) + 4\sin(t) + 4
cos(t) - 4\cos(t)\sin(t) \, dt.  $$
