<s>[INST] Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
 [/INST] $z=(-x-y)$ leads to
$$ x^2+y^2+(x+y)^2 = 6 $$
that is equivalent to $x^2+xy+y^2=3$, i.e. the equation of an ellipse in the $xy$-plane. Such ellipse goes through $(1,1)$, hence by considering the intersections between such ellipse and the lines through $(1,1)$, i.e. by solving $x^2+xy+y^2=3$ and $y=m(x-1)+1$, we get that
$$ (x,y)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1}\right) $$
for $m\in\overline{\mathbb{R}}$ is a parametrization of the ellipse in the $xy$-plane and 
$$ (x,y,z)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1},\frac{m^2+4m+1}{m^2+m+1}\right) $$
is a parametrization of the wanted section (circle) in the Euclidean space.
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