To evaluate the line integral of the vector field $\mathbf{F} = \langle 3xy, -xz, 2 \rangle$ along the curve $C$ formed by traveling along the unit circle from $(0, 1, 1)$ to $(1, 0, 1)$ and then traveling along the straight line from $(1, 0, 1)$ to $(2, 1, 1)$, we can split the problem into two parts corresponding to the two segments of the curve and then sum the results.

1. Traveling along the unit circle from $(0, 1, 1)$ to $(1, 0, 1)$

Parametrize the unit circle segment in the $xy$-plane while $z = 1$: $\mathbf{r}_1(t) = (\cos t, \sin t, 1), \quad t \in \left[ \frac{\pi}{2}, 0 \right]$

The differential $d\mathbf{r}_1 = \frac{d\mathbf{r}_1}{dt} dt = (-\sin t, \cos t, 0) dt$.

Evaluate the integral: $\int_{C_1} 3xy \, dx - xz \, dy + 2 \, dz = \int_{\frac{\pi}{2}}^{0} \left( 3 (\cos t)(\sin t)(-\sin t) - (\cos t)(1)(\cos t)(\cos t) + 2 \cdot 0 \right) dt$

Simplify the integrand: $= \int_{\frac{\pi}{2}}^{0} \left( -3\cos t \sin^2 t - \cos^3 t \right) dt$

$= \int_{\frac{\pi}{2}}^{0} \left( -3\cos t (1 - \cos^2 t) - \cos^3 t \right) dt$

$= \int_{\frac{\pi}{2}}^{0} \left( -3\cos t + 3\cos^3 t - \cos^3 t \right) dt$

$= \int_{\frac{\pi}{2}}^{0} \left( -3\cos t + 2\cos^3 t \right) dt$

Evaluate this integral by parts or appropriate substitutions: $\int_{\frac{\pi}{2}}^{0} -3\cos t \, dt + \int_{\frac{\pi}{2}}^{0} 2\cos^3 t \, dt$

The first integral is straightforward: [ -3 \int_{\frac{\pi}{2}}^{0} \cos t , dt = -3 \sin t