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: $\gamma(t)=(\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,-\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,\frac{\sqrt{6}}{3}sint) , \,\,\,\, t \in [0,2\pi]$

Hint
Let $D=\frac{|d|}{\sqrt{a^2+b^2+c^2}}$ be the distance between the plane and the center of the sphere.
There  are three cases
$D>a \implies $ no interesection
$D=a \implies $the plane is tangent.
$D<a \implies $ the interesection is a circle  of radius$=\sqrt{a^2-D^2}$.
