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?

The equation
$$
x^2 + y^2 + z^2 = 6
$$
gives a sphere around the origin with radius $R=\sqrt{6}$.
$$
0 = x + y + z = (1,1,1) \cdot (x,y,z)
$$
gives a plane $H$ with normal vector $(1,1,1)^\top$ including the origin.
The intersection is a circle of radius $R$ on that plane $H$, with the origin as midpoint.
GeoGebra seems to give:
$$
u(t) = (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t))^\top
$$
Checking:
$$
(-\sqrt{3} \cos(t) +\sin(t))^2 + (\sqrt{3} \cos(t) + \sin(t))^2 + (-2 \sin(t))^2 = \\
3 \cos^2(t) + \sin^2(t) - 2\sqrt{3} \cos(t)\sin(t) + \\
3 \cos^2(t) + \sin^2(t) + 2\sqrt{3} \cos(t)\sin(t) + \\
4 \sin^2(t) = \\
6 \cos^2(t) + 6 \sin^2(t) = 6
$$
so the curve lies on the sphere.
$$
(1,1,1) \cdot (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t)) = 
0
$$
so that curve is on the plane $H$.
How to get there?
One might try a coordinate transformation of a circle in the $x$-$y$-plane with radius $r$ onto that plane by two rotations:
We would need
$$
T\,(0,0,1)^\top = (1,1,1)/\sqrt{3}
$$
