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: $\gamma(t)=(\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,-\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,\frac{\sqrt{6}}{3}sint) , \,\,\,\, t \in [0,2\pi]$
 A: 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:
$$
{\bf p} \cdot {\bf n} = \delta \,\quad {\bf p} \cdot {\bf p} = r^{\,2} 
$$
A: 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.
A: 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$=\sqrt{a^2-D^2}$.
