Pairwise-intersecting circles $R,G,B$ have concurrent common chords?

The equations of the three circles are

$ (x - x_1)^2 + (y - y_1)^2 = r_1^2 \hspace{30pt}(1) $

$ (x - x_2)^2 + (y - y_2)^2 = r_2^2 \hspace{30pt}(2)$

$ (x - x_3)^2 + (y - y_3)^2 = r_3^2 \hspace{30pt}(3)$

Intersecting $(1)$ with $(2)$, the intersection points lie on the line

$ - 2 x (x_1 - x_2) - 2 y (y_1 - y_2) = r_1 ^2 - r_2 ^ 2 \hspace{30pt}(4)$

Intersecting $(1)$ with $(3)$, the intersection points lie on the line

$ - 2 x (x_1 - x_3) - 2 y( y_1 - y_3) = r_1^2 - r_3^2 \hspace{30pt} (5)$

And finally interecting $(2)$ with $(3)$, the intersection points lie on the line

$ - 2 x (x_2 - x_3) - 2 y (y_2 - y_3) = r_2 ^ 2 - r_3 ^ 2 \hspace{30pt}(6)$

If $(x, y)$ satisfies $(4)$ and $(5)$ it must satisfy $(6)$. This can seen by subtracting equation $(4)$ from $(5)$.

Therefore, yes, the three line segments always meet at a single point.

Call $p$ the intersection of $uw$ and $xy$. The line $zp$ intersects $G$ in $s_1$ and $B$ in $s_2$.

Using the Intersecting Chords Theorem we have

\begin{align} pz\cdot ps_1 & = px\cdot py &\text {(power of $p$ in $G$)} \\ & = pu\cdot pw &\text {(power of $p$ in $R$)} \\ & = pz\cdot ps_2 &\text {(power of $p$ in $B$)} \\ \end{align}

Then $ps_1=ps_2$ and hence $s_1=s_2=s$