How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$ and find it arc length? How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$?
Should I make a new equation $x^{2} + y^{2} + z^{2} = x + y + z$?
Or do I need to use spherical coordinates to parametrize $x^{2} + y^{2} + z^{2} = 1$ ?
Also, I want to find the arc length of this parametric equation.

Yet another answer.
The unit sphere in $\mathbb{R}^n$ can be described by $\|x\|^2 = 1$, and the equivalent plane by $e^T x = 1$ where $e=(1,...,1)$. Let $Q$ be an orthogonal rotation such that $Qe = \sqrt{n}e_1$, where $e_1 = (1,0,...,0)$.
Using the transformation $y=Qx$ we get the equivalent equations $\|y\|^2 = 1$ and $e_1^Ty = {1 \over \sqrt{n}}$. Equivalently
$y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$, so we see that this is an $n-1$ dimensional $\sqrt{1-{1 \over n}}$ sphere centered at $({1 \over \sqrt{n}},0,...,0)$.
Mapping back we see that the intersection is parameterised by
${1 \over n} e + \sum_{k=2}^n y_kQ^T e_k$, where $y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$.
For $n=3$ we can parameterise this as
${1 \over 3}(1,1,1) + {\sqrt{2 \over 3}}\cos \theta \ u +{\sqrt{2 \over 3}}\sin \theta \ v$ where ${1 \over \sqrt{3}} (1,1,1),u,v$ are orthonormal.
