Parametrization of the intersection between a sphere and a plane I can't find a way to get the parametric equation $\gamma(t)=(x(t),y(t),z(t))$ of a curve that is the intersection of a sphere and a plane (not parallel to any coordinate planes).  That is
$$\begin{cases} x^2+y^2+z^2=r^2 \\ ax+by+cz=d \end{cases}$$
I don't know how to move the variables in order to get something easily parameterizable.\
Can anyone let me know the main steps to get the parameterization of this type of curve?

Example
$$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 \end{cases}$$
Answer: $\gamma(t)=(\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,-\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,\frac{\sqrt{6}}{3}sint) , \,\,\,\, t \in [0,2\pi]$

If the plane passes through the origin (so $d = 0)$, the intersection will be a circle of radius $1$ that lies on the plane. Thus, you can choose a pair $(e_1,e_2)$ of orthogonal vectors of unit length that lie on your plane and then the intersection will be parametrized as
$$ \gamma(t) = \cos(t) e_1 + \sin(t) e_2. $$
To calculate this explicitly, you can choose $e_1$ to be any unit length vector that lies on the plane and set $e_2 = e_1 \times \frac{(a,b,c)}{\|(a,b,c)\|}$ (as $(a,b,c)$ is the normal vector to the plane). For example, for $x + y + z = 0$ we can take
$$ e_1 = \frac{(1,-1,0)}{\|(1,-1,0)\|} = \frac{1}{\sqrt{2}}(1,-1,0), \\
e_2 = \frac{1}{\sqrt{2}} (1,-1,0) \times \frac{1}{\sqrt{3}} (1,1,1) = \frac{1}{\sqrt{6}} (-1,-1,2) $$ 
and so
$$ \gamma(t) = \cos(t) \frac{1}{\sqrt{2}} (1,-1,0) + \sin(t) \frac{1}{\sqrt{6}} (-1,-1,2) = \left( \frac{\cos t}{\sqrt{2}} - \frac{\sin t}{\sqrt{6}}, -\frac{\cos t}{\sqrt{2}} - \frac{\sin t}{\sqrt{6}}, \frac{2}{\sqrt{6}} \sin t\right) $$
which is almost the parametrization up wrote but note that you have a sign error in your parametrization.
