According to the question, there are two blocks of masses $m_1$ and $m_2$ which are kept in contact on inclined plane of inclination $60^{\circ}$ as shown in the figure,


Coefficient of friction between the surface of plane and surface of first block, $\mu_1=1.5 \mu$
Coefficient of friction between the surface of plane and surface of second block, $\mu_2=1 \mu$
Now, free body diagram for the first block,


If $a$ be the common acceleration, then
$$
\begin{aligned}
& m_1 g \sin 60^{\circ}-\mu_1 \cdot R_1+R=m_1 a \\
& m_1 g \frac{\sqrt{3}}{2}-1.5 \mu m_1 g \cos 60^{\circ}+R=m_1 a \\
& m_1 g \frac{\sqrt{3}}{2}-\frac{1.5 \mu m_1 g}{2}+R=m_1 a
\end{aligned}
$$
Similarly, for second body,


$$
\begin{gathered}
m_2 g \sin 60^{\circ}-\mu_2 R_2-R=m_2 a \\
m_2 g \frac{\sqrt{3}}{2}-\mu m_2 g \cos 60^{\circ}-R=m_2 a \\
m_2 g \frac{\sqrt{3}}{2}-\frac{\mu m_2 g}{2}-R=m_2 a
\end{gathered}
$$
From Eqs. (i) and (ii), we get
$$
\frac{m_1 g \frac{\sqrt{3}}{2}-\frac{1.5 \mu m_1 g}{2}+R}{m_2 g \frac{\sqrt{3}}{2}-\frac{\mu m_2 g}{2}-R}=\frac{m_1 a}{m_2 a}
$$

$$
\begin{aligned}
& \Rightarrow \quad \frac{\sqrt{3}}{2} m_1 m_2 g-0.75 \mu m_1 m_2 g+m_2 R \\
& =\frac{\sqrt{3}}{2} m_1 m_2 g-0.5 \mu m_1 m_2 g-m_1 R \\
& \Rightarrow\left(m_1+m_2\right) R=0.25 \mu m_1 m_2 g \\
& \Rightarrow \quad R=\frac{\mu m_1 m_2 g}{4\left(m_1+m_2\right)} \\
&
\end{aligned}
$$
Hence, the force of reaction between the blocks doing the motion is, $R=\frac{1}{4} \frac{m_1 m_2}{\left(m_1+m_2\right)} \mu g$.