Let $F=(-2y,-z,2x)$.
Let $\partial D$ be a boundary of the intersection of: $$D=\{x,y,z: x+y+z=2\}\cap\{x,y,z:x^2+y^2+z^2=4 \}$$
Need to find $$\oint_{\partial D}\vec{F}$$ Naturally, this problem is solved by Stokes theorem, but I wanted to do so with standard parameterization.
$$ x=2,y=0,z=0 \\ x=0,y=2, z=0 \\ x=0,y=0,z=2$$ These three points exist on said circle, so we can find it's centre.
Let $a,b,c$ be the centre of said circle, then $$ (2-a)^2+(0-b)^2+(0-c)^2= \\(0-a)^2+(2-b)^2+(0-c)^2=\\(0-a)^2+(0-b)^2+(2-c)^2$$ Solving this gives us $a=b=c$ After putting $a+a+a=2$ we find that the circle centre is $(\frac{2}{3},\frac{2}{3},\frac{2}{3})$
We can easily find the radius now.
Now let's find the parameterization for the circle with $$\partial D(t)= (x(t),y(y),z(t))$$
In 2d, such parameterization would be $(a+r\cos(t),b+r\sin(t))$
According to a video on youtube, this time the parameterization will be: $$ x(t)=c_1+a_1r\cos(t)+b_1r\sin(t)\\ y(t)=c_2+a_2r\cos(t)+b_2r\sin(t)\\ z(t)=c_3+a_3r\cos(t)+b_3r\sin(t)$$ Where $(c_1,c_2,c_3)$ is the circle centre and $A,B$ are some projection vectors.
I tried finding them but couldn't figure how.
Assuming $(2,0,0) $ is the starting location for our integral and $t\in [ 0,2\pi]$ then we get $$ 2= \frac{2}{3} + a_1 r \\ 0=\frac{2}{3} + a_2 r \\ 0= \frac{2}{3} + a_3 r $$
Now I can find $A$ but not sure how to obtain B, as I can't just use any point. Would appreciate guidance to find that vector, or how to find $ A,B$ systematically and not just putting points.