Find the parametric equation

First, what is a parametrization of the intersection of the sphere $x^2 + y^2 + z^2 = a^2$ and the plane $z = 1/\sqrt{3}$? This is easy: $(x(\theta), y(\theta), z(\theta)) = \left( \sqrt{a^2 - 1/3} \cos \theta, \sqrt{a^2 - 1/3} \sin \theta, 1/\sqrt{3} \right),$ which, as we expect, is defined if and only if $a \ge 1/\sqrt{3}$. Now all we need to do is rotate this parametrization in the plane $x = y$ by a suitable angle $\alpha = \tan^{-1} \sqrt{2}$. But it's not quite that simple; we multiply by an initial rotation matrix $R_{xz}(\alpha) = \begin{bmatrix} \cos\alpha & 0 & \sin\alpha \\ 0 & 1 & 0 \\ -\sin\alpha & 0 & \cos \alpha \end{bmatrix},$ followed by a (counterclockwise) rotation of $\beta = \pi/4$ in the $xy$-plane: $R_{xy}(\beta) = \begin{bmatrix} \cos \beta & -\sin\beta & 0 \\ \sin\beta & \cos\beta & 0 \\ 0 & 0 & 1 \end{bmatrix}.$ Therefore the transformed parametrization is $(x'(\theta), y'(\theta), z'(\theta)) = R_{xy}(\pi/4) R_{xy}(\tan^{-1}\sqrt{2}) \cdot (x(\theta), y(\theta), z(\theta)),$ which gives $\begin{aligned} x'(\theta) &= \frac{1}{3}\left( 1 + \sqrt{6a^2-2} \cos(\theta+\frac{\pi}{3}) \right), \\ y'(\theta) &= \frac{1}{3} \left(1 + \sqrt{6a^2-2} \cos(\theta-\frac{\pi}{3}) \right), \\ z'(\theta) &= \frac{1}{3} \left( 1 - \sqrt{6a^2-2} \cos\theta \right), \end{aligned}$ for $\theta \in [0, 2\pi)$. It is then easy to verify that $x' + y' + z' = 1$ and $x'^2 + y'^2 + z'^2 = a^2$. Note that any such parametrization is not unique: we can find any number of equivalent parametrizations for this curve.