Since the free body diagram is as follows:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/y5T75.png)
the four reactions can be calculated by imposing the respective equilibrium on the translation:
$$
\begin{cases}
R_A - \frac{1}{2}R_B = 0 \\
\frac{\sqrt{3}}{2}R_B - P_1 = 0 \\
\frac{1}{2}R_B - R_D = 0 \\
-\frac{\sqrt{3}}{2}R_B + R_C - P_2 = 0 \\
\end{cases}
\quad \quad \Leftrightarrow \quad \quad
\begin{cases}
R_A = \frac{\sqrt{3}}{3}P_1 \\
R_B = \frac{2\sqrt{3}}{3}P_1 \\
R_C = P_1 + P_2 \\
R_D = \frac{\sqrt{3}}{3}P_1 \\
\end{cases}\,.
$$
If for verification we want to impose equilibrium on the rotation with respect to $B$:
$$
-R_A\frac{\sqrt{3}}{2}r + P_1\frac{1}{2}r + R_C\frac{1}{2}r - R_D\frac{\sqrt{3}}{2}r - P_2\frac{1}{2}r = 0
$$
i.e.
$$
-\frac{\sqrt{3}}{3}P_1\frac{\sqrt{3}}{2}r + P_1\frac{1}{2}r + (P_1+P_2)\frac{1}{2}r - \frac{\sqrt{3}}{3}P_1\frac{\sqrt{3}}{2}r - P_2\frac{1}{2}r = 0
$$
which result in an identity $0=0$, the check is satisfied.
By imposing equilibrium on the translation and rotation of the system of the two balls:
$$
\begin{cases}
R_A - R_D = 0 \\
R_C - P_1 - P_2 = 0 \\
-R_A\sqrt{3}\,r - R_C\,r + P_1\,2\,r + P_2\,r = 0
\end{cases}
\quad \quad \Leftrightarrow \quad \quad
\begin{cases}
R_A = \frac{\sqrt{3}}{3}P_1 \\
R_C = P_1 + P_2 \\
R_D = \frac{\sqrt{3}}{3}P_1 \\
\end{cases}\,.
$$
Naturally, if $R_B$ is needed, the body system must be exploded as above.