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

Just to present a vectorial approach to the problem

Consider the plane equation written as:
$$
\frac{{a\,x + b\,y + c\,z}}{{\sqrt {a^{\,2}  + b^{\,2}  + c^{\,2} } }} = \frac{d}{{\sqrt {a^{\,2}  + b^{\,2}  + c^{\,2} } }}
$$
That means that the points ${\bf p} = \left( {x,y,z} \right)$ on the plane
shall project onto the unit vector ${\bf n} = \left( {a,b,c} \right)/\sqrt {a^{\,2}  + b^{\,2}  + c^{\,2} } $
at a constant distance $\delta  = d/\sqrt {a^{\,2}  + b^{\,2}  + c^{\,2} } $, i.e.
$$
{\bf p} \cdot {\bf n} = \delta 
$$
so that the plane is distant $\delta$ from the origin.
At the same time we shall have 
$$
\left| {\,{\bf p}\,} \right| = r
$$
So with reference to the sketch, we can put
$$
{\bf p} = \delta \,{\bf n} + \sqrt {r^{\,2}  - \delta ^{\,2} } \,{\bf t} = \delta \,{\bf n} + \rho \,{\bf t}
$$
where ${\bf t}$  is a generic unit vector parallel to the plane, that is normal to ${\bf n}$.
We can express ${\bf t}$ by taking two unit vectors ${\bf u}$ and ${\bf v}$, normal to ${\bf n}$ and to each other,
and then putting 
$$
{\bf t} = \cos \theta \,{\bf u} + \sin \theta \,{\bf v}
$$
To determine ${\bf u}$ and ${\bf v}$ we can take (if, e.g. $c \ne 0$)
$$
{\bf u} = \left( {0, - c,b} \right)/\sqrt {b^{\,2}  + c^{\,2} } \quad \quad {\bf v} = {\bf n} \times {\bf u}
$$
example
with $r=4 \; a=1 \; b=2 \; c=3 \; d=7$ we get
$$
{\bf n} = \,\sqrt {14} /14\,\left( {\begin{array}{*{20}c}
   1  \\
   2  \\
   3  \\
\end{array}} \right)\;\quad \delta  = \sqrt {14} /2\quad \rho  = 5\sqrt 2 /2
$$
$$
{\bf u} = \sqrt {13} /13\;\left( {\begin{array}{*{20}c}
   1  \\
   2  \\
   3  \\
\end{array}} \right)\quad {\bf v} = \sqrt {182} /182\;\left( {\begin{array}{*{20}c}
   {13}  \\
   { - 2}  \\
   { - 3}  \\
\end{array}} \right)
$$
and thus
$$
\begin{array}{l}
 {\bf p} = \delta \,{\bf n} + \rho \,\left( {\cos \theta \,{\bf u} + \sin \theta \,{\bf v}} \right)\quad  \Rightarrow  \\ 
  \Rightarrow \quad \left( {\begin{array}{*{20}c}
   x  \\
   y  \\
   z  \\
\end{array}} \right) = \frac{1}{2}\,\left( {\begin{array}{*{20}c}
   1  \\
   2  \\
   3  \\
\end{array}} \right) + \frac{{5\sqrt {364} }}{{364}}\,\left( {\sqrt {14} \cos \theta \;\left( {\begin{array}{*{20}c}
   1  \\
   2  \\
   3  \\
\end{array}} \right) + \sin \theta \,\;\left( {\begin{array}{*{20}c}
   {13}  \\
   { - 2}  \\
   { - 3}  \\
\end{array}} \right)} \right) \\ 
 \end{array}
$$
and you can verify that:
$$
{\bf p} \cdot {\bf n} = \delta \,\quad {\bf p} \cdot {\bf p} = r^{\,2} 
$$
