curve on sphere from $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$ Consider the set:
$$
A \equiv \left\{ (x,y,z) \in \mathbb{R}^{3} | x + y + z = 0\ \mathrm{and}\ x^{2} + y^{2} + z^{2} = 1 \right\}
$$
This is the intersection of the unit sphere with the above plane.
Is there a way to parametrize this set with a curve $\gamma : [0,1] \to A$ going from the point $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$?
I simply can't figure it out since we have the extra constraint of being in the plane $x+y+z=0$

When a plane intersects a sphere the curve of intersection will be a circle.
the origin is at the center of this cricle (because the plane goes through the origin, and the sphere is centered at the origin).  Now you just need to parametarize the circle.
We have two points on this circle, but what we really need are orthogonal vectors.  We can use one of the points, and find a vector in the plane that is orthogonal.
$(-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)$ is a vector orthogonal to $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ that lies in the plane.
$(x,y,z) = (-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cos t + (-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)\sin t$
Will trace out the circle.
And for what bounds of $t$ will will that equation trace to the endpoints we seek?
$(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cdot(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}) = 0.5$
$t = \frac{\pi}{3}$ or $\frac{-\pi}{3}$
$t\in[0,\frac {\pi}{3}]$ 
