Parametric equations for intersection between plane and circle So I was looking at this question Determine Circle of Intersection of Plane and Sphere
but I need to know how to find a parametric equation for intersections such as these.  My particular question is to find a parametric equation for the intersection between the plane 
$$x+y+z=1$$ and unit sphere centered at the origin.  
I started out my question by substituting 
$$
z=-x-y+1$$ into 
$$x^2+y^2+z^2=1  $$ deriving 
$$x^2+y^2+(-x-y+1)^2=1$$
and getting $$2x^2+2y^2+2xy-2x-2y=0$$ but I am unsure how to proceed from here.
I also tried to use the vector equation of the plane 
$$r(u, v)=(0,0,1)+(2,1,-3)u+(1,1,-2)v$$ but I am not sure how that would help.

The plane $ \zeta = \{(x,y,z) : x + y + z = 1\},$ is perpendicular to the vector
$\langle 1,1,1 \rangle$.
Two points on $\zeta$ are 
$\mathbf O = \left( \dfrac 13,\ \dfrac 13,\ \dfrac 13 \right)
\text{ and }
\mathbf A = (0,0,1)$.
Since $\overrightarrow{\mathbf{OA}} = \dfrac 13 \langle -1, -1, 2 \rangle$,
 the unit vector
$\overrightarrow u = \dfrac {1}{\sqrt 6} \langle -1, -1, 2 \rangle$ 
is parallel to the plane $\zeta$.
Since $\overrightarrow{\mathbf{OA}} \times
 \langle 1,1,1 \rangle = \langle -1, 1, 0 \rangle$, then the unit vector 
$\overrightarrow v = \dfrac{1}{\sqrt 2} \langle -1, 1, 0 \rangle$ is parallel to the plane $\zeta$ and 
$\overrightarrow u \perp \overrightarrow v$.
So we can parameterize the plane, $\zeta$ as
$$\zeta(s,t)
  = \mathbf O + s \overrightarrow u + t \overrightarrow v
  = \left(
         \dfrac 13 -  \dfrac{s}{\sqrt 6} - \dfrac{t}{\sqrt 2},
         \dfrac 13 -  \dfrac{s}{\sqrt 6} + \dfrac{t}{\sqrt 2},
         \dfrac 13 + \dfrac{2s}{\sqrt 6}
    \right)$$
As ugly as the following looks
$$\left( \dfrac 13 -  \dfrac{s}{\sqrt 6} - \dfrac{t}{\sqrt 2}\right)^2 +
\left( \dfrac 13 -  \dfrac{s}{\sqrt 6} + \dfrac{t}{\sqrt 2}\right)^2 +
\left( \dfrac 13 +  \dfrac{2s}{\sqrt 6}\right)^2 = 1 $$
It simplifies to $s^2 + t^2 = \dfrac  23$
So we let
$s = \sqrt{\dfrac 23}\cos \theta$ and 
$t = \sqrt{\dfrac 23}\sin \theta$ and simplify. We get
$$\zeta(s,t)
  = \left(
         \dfrac 13 -  \dfrac {\cos \theta}{3}  - \dfrac{\sin \theta}{\sqrt 3},\;
         \dfrac 13 -  \dfrac {\cos \theta}{3} + \dfrac{\sin \theta}{\sqrt 3},\;
         \dfrac 13 +  \dfrac {2\cos \theta}{3}
    \right)$$
a more intuitive answer

Let $S$ be the unit sphere $x^2 + y^2 + z^2 = 1$.
Let $P$ be the plane $x + y + z = 1$.
Let $C$ be the circle $C = S \cap P$.
The unit vector $U = \dfrac{1}{\sqrt 3} \langle 1,1,1 \rangle$ is perpendicular to $P$.
The line $x = y = z$


*

*passes through, $(0,0,0)$,  the center of $S$

*passes through, $X$, the center of $C$

*is perpendicular to $P$
The distance from $P$ to the center of $S$ is
$ \dfrac{\left| 0 + 0 + 0 - 1 \right|}{\sqrt{1^2 + 1^2 + 1^2}} 
   = \dfrac{1}{\sqrt 3}$.
The center of $C$ is at
    $X = (0,0,0) + \dfrac{1}{\sqrt 3}U
= \left( \dfrac 13,\dfrac 13, \dfrac 13 \right)$.
The radius of $C$ is $r = \sqrt{1 - \dfrac 13} = \sqrt{\dfrac 23}$.
We need to find two points, $A$ and $B$, on $C$ such that
$\overrightarrow{XA} \perp \overrightarrow{XB}$.


*

*$A = (1,0,0)$ is a point on $C$.

*$\left \| \overrightarrow{XA} \right \|  =
   \left \| \left( \dfrac 23, -\dfrac 13, -\dfrac 13  \right) \right \| =
   \sqrt{\dfrac 23}$

*$U \times \sqrt{\dfrac 32} \; \overrightarrow{XA} = 
\left( 0, \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2}  \right)$
is a unit vector, in P, that is perpendicular to $\overrightarrow{XA}$.

*$\overrightarrow{XB} = 
\sqrt{\dfrac 23} \left( 0, \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2}  \right) =
\left( 0, \dfrac{1}{\sqrt 3}, -\dfrac{1}{\sqrt 3}  \right)$
\begin{align}
C &=
  X 
  + \cos \theta \; \overrightarrow{XA}
  + \sin \theta \; \overrightarrow{XB}
\\
C &= \left( \dfrac 13,\dfrac 13, \dfrac 13 \right) 
  + \cos \theta \; \left( \dfrac 23, -\dfrac 13, -\dfrac 13  \right)
  + \sin \theta \; \left( 0, \dfrac{\sqrt 3}{3}, -\dfrac{\sqrt 3}{3}  \right) 
\\
C &= \dfrac 13(1 + 2 \cos \theta, \;
               1 - \cos \theta + \sqrt 3 \sin \theta, \;
               1 - \cos \theta - \sqrt 3 \sin \theta)
\end{align}
