For any point $P$ on unit sphere, let $P'$ be its antipodal point and $\hat{P}$ the corresponding unit vector. We will use the notation
$\mathcal{C}_{P_1P_2\cdots P_n}$ to denote a spherical arc starting from $P_1$, passing through $P_2,\ldots$ and end at $P_n$.
Let $A, B, C$ be any three points on unit sphere, close enough to fit
within half of a hemisphere
(i.e. a spherical lune of angle $\frac{\pi}{2}$ ).
Let $\Omega_{ABC}$ be the area of spherical $\triangle ABC$.
It can be computed using a formula by Oosterom and Strackee
$$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{ \left|\hat{A}\cdot (\hat{B} \times \hat{C})\right|}{1 + \hat{A}\cdot\hat{B} + \hat{B}\cdot\hat{C} + \hat{C}\cdot\hat{A}}$$
In the special case where $A, B$ lies on the equator, symmetric with respect to $x$-axis and $C$ lies on the upper hemisphere, i.e.
$$
\begin{cases}
A &= (\cos\alpha,-\sin\alpha,0),\\
B &= (\cos\alpha,+\sin\alpha,0),\\
C &= (x, y, z)\end{cases}
\quad\text{ where } \alpha \in (0,\frac{\pi}{2}), z > 0$$
Above formula reduces to
$$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{\sin\alpha z}{\cos\alpha + x}$$
This implies the locus of $P$ in upper hemisphere for fixed
$\Omega_{ABP} = \Omega_{ABC}$ is the circular arc $\mathcal{C}_{B'CA'}$.
Let $\theta$ be the angle between $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$ at $B'$. The plane holding the locus has the form
$$t ( c + x ) - s z = 0\quad\text{ where }\quad
\begin{cases}
t &= \tan\left(\frac{\Omega_{ABC}}{2}\right)\\
c &= \cos\alpha\\
s &= \sin\alpha
\end{cases}$$
The normal vector of the plane is pointing along the direction $(t, 0, -s)$. This means the tangent vector of $\mathcal{C}_{B'CA'}$ at $B'$ is along the direction $(t, 0, -s ) \times ( -c, -s, 0 ) \propto (s, -c, t)$.
Notice the tangent vector of $\mathcal{C}_{B'ABA'}$ at $B'$
is pointing along the direction $(s,-c,0)$, we find
$$\cos\theta = \frac{s^2 + c^2 + 0}{\sqrt{s^2+c^2}\sqrt{s^2+c^2+t^2}} = \cos\left(\frac{\Omega_{ABC}}{2}\right)$$
From this, we can deduce the circular arcs
$\mathcal{C}_{B'PA'}$ and $\mathcal{C}_{B'ABA'}$ intersect at an angle $\frac{\Omega_{ABC}}{2}$.
This leads to following construction of the desired "spherical centroid" $X$.
Construct the circular arcs $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$,
Trisect the angle $\angle AB'C$ - i.e. find a circular arc
$\mathcal{B'\tilde{C}A'}$ such that $\tilde{C}$ is lying on same side as $C$ with respect to $AB$ and $\angle AB'\tilde{C} = \frac13 \angle AB'C$.
Repeat above procedures for other two sides of $\triangle ABC$ to get
circular arcs $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.
$X$ will be lying on the common intersection of the three circular arcs
$\mathcal{C}_{B'\tilde{C}A'}$, $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.
The equation of a variable plane in intercept form is $$\frac xa+\frac yb+ \frac zc=1$$
Here, $a,b$ and $c$ are intercept on $x,y$ and $z$ axes.
Given, it's distance from origin is constant $$\frac{\Big|\Big(\frac xa+\frac yb+ \frac zc-1\Big)_{x=y=z=0}\Big|}{\sqrt{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}}}=p(\text{constant})$$
And centroid of triangle $$(x,y,z)=\left(\frac a3,\frac b3,\frac c3\right)$$
Therefore you get $$\frac{1}{p^2}=\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}$$
Also, $a=3x,b=3y$ and $c=3z$
$$\implies \frac{1}{p^2}=\frac {1}{9x^2}+\frac {1}{9y^2}+\frac {1}{9z^2}$$
Hence the desired locus is :
$$\color{blue}{\frac {1}{x^2}+\frac {1}{y^2}+\frac {1}{z^2}=\frac{9}{p^2}}$$
Best Answer
The spherical centroid exists by the spherical version of Ceva's theorem.
Assuming that $A,B,C$ are three points on a unit sphere centered at $O$, we may join $A$ with the midpoint $M_A$ of the $BC$ side in the spherical triangle $ABC$. The plane through $A,M_A,O$ meets the $ABC$ plane at a line $\ell_A$, the plane through $B,M_B,O$ meets the $ABC$ plane at a line $\ell_B$. Assuming that $\ell_A$ and $\ell_B$ meet at $G$ in the $ABC$ plane, the spherical centroid of the spherical triangle $ABC$ is just the intersection between the $OG$ ray and the original sphere, i.e. $\frac{G}{\left\|G\right\|}$.
It follows that you just need to compute the (planar) centroid of the euclidean triangle $ABC$, since the $OM_A$ ray meets the $BC$ segment at its midpoint on so on.
Long story short, the answer is just given by $\color{red}{\frac{u+v+w}{\left\|u+v+w\right\|}}$, since the spherical medians are given by the central projections of the medians of the planar triangle $ABC$.