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?
 A: 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
$$
T\,(0,0,1)^\top = (1,1,1)/\sqrt{3}
$$
A: When you substitute $z=-(x+y)$ into $x^2+y^2+z^2=6$ we get 
\begin{eqnarray*}
x^2+y^2+(x+y)^2=6 \\
x^2+xy+y^2 =3.
\end{eqnarray*}
Now multiply this by $4$ & complete the square 
\begin{eqnarray*}
(2x+y)^2+3y^2=12.
\end{eqnarray*}
This is easily parameterised by $2x+y = \sqrt{12} \cos ( \theta)$ and $y = 2 \sin ( \theta)$.
\begin{eqnarray*}
x &=& \frac{\sqrt{12} \cos ( \theta)-2 \sin ( \theta)}{2} \\
y &=& 2 \sin ( \theta) \\
z &=& \frac{-\sqrt{12}  \cos ( \theta)- 2\sin ( \theta)}{2}
\end{eqnarray*}
A: $z=(-x-y)$ 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.
A: With spherical coordinate system, $r=\sqrt{6}$ and
$$z=-x-y=-\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi$$
so your parametric equation will be
$$
\begin{aligned}x&=\sqrt{6}\,\sin \theta \,\cos \varphi \\y&=\sqrt{6}\,\sin \theta \,\sin \varphi \\z&=\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi \end{aligned}$$
where $ θ ∈ [0, π], φ ∈ [0, 2π)$. 
