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

Step 1 ：First, we need to parameterize the curve C. Since it's a circle of radius 2 centered at (0,1,3) in the plane x+2y+3z=11, we can use the parameterization r(t) = (2cos(t), 1+2sin(t), 3-2sin(t)) for t in [0, 2π].

Step 2 ：Next, we compute the derivative of r(t) with respect to t, which gives us r'(t) = (-2sin(t), 2cos(t), -2cos(t)).

Step 3 ：Now, we substitute r(t) into the vector field F to get F(r(t)) = (6cos(t)-(1+2sin(t))^2, -4cos(t)sin(t)+3(3-2sin(t)), 3(1+2sin(t))+2).

Step 4 ：Then, we compute the dot product of F(r(t)) and r'(t), which gives us F(r(t))·r'(t) = -12cos(t)sin(t) + (1+2sin(t))^2sin(t) - 8cos^2(t)sin(t) + 6cos(t) - 6sin(t) + 6sin(t) + 2 = -12cos(t)sin(t) + sin(t) + 4sin^2(t)sin(t) - 8cos^2(t)sin(t) + 6cos(t) + 2.

Step 5 ：Finally, we compute the line integral ∫C F·dr = ∫ from 0 to 2π of F(r(t))·r'(t) dt. This is a standard one-dimensional integral that can be computed using techniques from calculus. The final result is ∫ from 0 to 2π of (-12cos(t)sin(t) + sin(t) + 4sin^2(t)sin(t) - 8cos^2(t)sin(t) + 6cos(t) + 2) dt = 2π.

Step 6 ：Therefore, the line integral ∫C F·dr is equal to \( \boxed{2π} \).