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?

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*}
