<s>[INST] 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.
 [/INST] First, let's name some points. Let $O=(0,0,0)$, $X=(1,0,0)$, $Y=(0,1,0)$, and $Z=(0,0,1)$. Let $A$ be the center of the circle we are trying to find. Let $K$ be the midpoint of $XY$. Note that $K=\left(\frac{1}{2},\frac{1}{2},0\right)$. 
Let's consider right triangle $ZOK$:

We note that A is the base of the altitude to the hypotenuse. Since $OZ=1$ and $OK=\frac{\sqrt{2}}{2}$, $OA=\frac{\sqrt{3}}{3}$. Because $A$ must be of the form $(n,n,n)$, we conclude that $A=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. We can also calculate that $AZ=\frac{\sqrt{6}}{3}$. Note that $AZ$ is the radius of the circle.

Now, let's invent a new 2D coordinate system with new unit vectors. $A$ will be the center of this coordinate system. Rotate $AZ$ 90 degrees clockwise on the plane $x+y+z=1$. This new vector (orange) will be one of our unit vectors, which we will call $u$. The other will be $AZ$ itself (pink), which we will call $v$. Imagine extending vector $u$ into a line and moving it down to the $xy$-plane. It makes a 45-45-90 triangle with the $x$ and $y$ axes. Since $u$ has no $z$-component, we conclude that $u=\left(-\frac{\sqrt{6}}{3}\cdot\frac{\sqrt{2}}{2},\frac{\sqrt{6}}{3}\cdot\frac{\sqrt{2}}{2},0\right)=\left(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},0\right)$. 

Drop a perpendicular from point A to the $z$ axis. Call the foot of this perpendicular $H$. Note that triangle $ZOA$ is similar to $ZOK$. Using this fact, we can calculate that $ZH=\frac{2}{3}$. Note that $ZH$ is the $z$-component of $v$. We can also calculate that $AH=\frac{\sqrt{2}}{3}$. Imagine moving $AH$ down to the $xy$-plane. We see that $AH$ forms 45 degrees with the $x$ and $y$ axes. We conclude that $v=\left(-\frac{\sqrt{2}}{3}\cdot\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{3}\cdot\frac{\sqrt{2}}{2},\frac{2}{3}\right)=\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)$.
The equation of the circle is $$A+u\cos(t)+v\sin(t)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)+\cos(t)\left(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},0\right)+\sin(t)\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)$$
Adding everything together, we get a final answer:
$$\frac{1}{3}\left(1-\sqrt{3}\cdot\cos(t)-\sin(t),1+\sqrt{3}\cdot\cos(t)-\sin(t),1+2\sin(t)\right)$$
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