Compute $\oint_{C} F \cdot d r$ where $\mathrm{C}$ is the circle of radius 2 centered at $(0,1,3)$ in the plane $x+2 y+3 z=11 . \quad F=\left(3 x-y^{2}\right) \uparrow+(-2 x y+3 z) \uparrow+(3 y+2) / T$ Compute $\int_{C} F \cdot d r$ where $C$ is any curve from $(1,0,9)$ to $(1,2,2)$.

Step 1 ：Parameterize the curve C as r(t) = (1, 2t, 9-7t), where t varies from 0 to 1.

Step 2 ：Compute the derivative of r(t) with respect to t to get the tangent vector to the curve at any point, dr/dt = (0, 2, -7).

Step 3 ：Substitute r(t) into the vector field F to get F(r(t)) = [-4t^2 + 3t, -4t^2 - 21t + 27, 6t + 2].

Step 4 ：Compute the dot product of F(r(t)) and dr/dt to get -8t^2 - 84t + 40.

Step 5 ：Integrate the dot product from t=0 to t=1 to get the line integral of F along C, which is -14/3.

Step 6 ：\(\boxed{-\frac{14}{3}}\) is the final answer.