I can only guess what you are asking. I, therefore, explain the whole thing in 2 steps.
I. Find the co-ordinates of the centroid, P(x, y) for ⊿ABC where $A= (x_1, y_1), B= (x_2, y_2)$ and $C=(x_3, y_3)$.
1.1 Let $D(p,q)$ be the midpoint of BC. Then, by mid-pt formula, $p = \frac {x_2 + x_3}{ 2}$, (similar result for q).
1.2 From geometrical knowledge, P(x, y) must lie on the median AD and, AP : PD = 2 : 1.
1.3 Apply the section formula to get $x = \frac {x_1 + x_2 + x_3} {3}$, (similar result for y).
II. Find $x_3$ if $x_1, x_2$ and $x$ are known.
2.1 Just plug in the given data at the right places in the developed formula, $x_3$ can then be found.
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 $ABC$.
Best Answer
Given three points in the coordinate plane $p_1 = (x_1,y_1)$, $p_2=(x_2,y_2)$, and $p_3=(x_3,y_3)$, the coordinates of the centroid $q$ is simply the average of the coordinates of the three points (actually, this is sometimes how the centroid is defined):
$$q = \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$$