To make things simpler I assume $\alpha_0=0$ (the desired value of $\alpha_0$ can always be added/subtracted at the end); furthermore I take the equator at $\delta=0$, so the range of the latitude $\delta$ is the interval $[-\pi/2,\pi/2]$. Your change of coordinates amounts to changing the standard basis $(e_1,e_, e_3)$ of ${\mathbb R}^3$ to the new basis
$$\bar e_1=(\cos\delta_0,0,\sin\delta_0), \quad \bar e_2=(0,1,0),\quad \bar e_3=(-\sin\delta_0,0,\cos\delta_0)\ .$$
It follows that the new coordinates $\bar x_k$ are given in terms of the old ones $x_i$ by the formulas
$$\bar x_1=\cos\delta_0 x_1 +\sin\delta_0 x_3, \quad \bar x_2=x_2,\quad \bar x_3=-\sin\delta_0 x_1+\cos\delta_0 x_3\ .$$
Now we have to express this in terms of the "geographical" quantities $\alpha$, $\delta$, resp. $\bar\alpha$, $\bar\delta$. On the one hand we have
$$x_1=\cos\delta\cos\alpha,\quad x_2=\cos\delta\sin\alpha, \quad x_3=\sin\delta\ ,$$
and on the other hand
$$\bar\alpha=\arg\Bigl({\bar x_1\over\rho},{\bar x_2\over\rho}\Bigr), \quad \bar\delta=\arcsin(\bar x_3)\ ,$$
where $\rho:=\sqrt{\bar x_1^2+\bar x_2^2}$. Putting it all together some simplifications will result.
Well, I ended up solving this myself after a little more thought. The solution is to convert from the (r, phi, theta) of spherical coordinates to the (x, y, z) of Cartesian coordinates, then swap the z-axis with one of the others before finally converting back to spherical coordinates again.
Best Answer
The parallel of latitude is actually a circle of radius $r\cos(\alpha)$
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Thus, the length of the parallel of latitude $\alpha$ is $2\pi r\cos(\alpha)$, where $r$ is the radius of the Earth.