<s>[INST] Integrating $F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}$ along a circle 
Let $F:\mathbb R^3 - \{0\}\to \mathbb R$ be given by $$F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}.$$ Compute $\int_C F\cdot ds$ where $C$ is the unit cirlce on the plane $x+y+z=3$ with the orientation from the point $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$ to $$\left(1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ to $$\left(1 + \frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ and back to $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$

What I've done: By Green's theorem, the integral is $$\iint_D \operatorname{curl}(F)\cdot k\ dA$$ I got that the curl is $$\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}}$$ so the integral is $$\int\int_D \frac{z}{(x^2+y^2+z^2)^{3/2}} \ dA$$ But I'm having difficulties with determining how to define the region $D$ to compute the last integral.
 [/INST] Obviously $D$ is a disk of radius $1$ in the given plane. You just need to find it's origin. The format of the given points on the circle hints that the center is $(1,1,1)$, which you can verify.
I would do a change of variables, to some $X,Y,Z$ such that $X,Y$ are in the plane, then $Z=0$. 
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