To solve this problem, we need to analyze the forces acting on the system.$\\$Given:$\\$• $m_1 = m_2 = m_3 = m \\$• Coefficient of friction $\mu \\$Let's denote:$\\$• $T$ as the tension in the string.$\\$• $a$ as the acceleration of the masses.$\\$For $m_1$ and $m_2$ on the table:$\\$The frictional force $f = \mu m g \\$Applying Newton's second law for $m_1$ and $m_2: \\ T - \mu m g = m a \quad$(1)$\\$For $m_3$ hanging freely:$\\ m g - T = m a \quad$(2)$\\$Add equations (1) and (2):$\\ (T - \mu m g) + (m g - T) = m a + m a \\ m g - \mu m g = 2 m a \\$Simplify:$\\ m g (1 - \mu) = 2 m a \\$Solve for $a: \\ a = \frac{g (1 - \mu)}{2} \\$Since $m_1 = m_2 = m_3 = m,$ the downward acceleration of $m_3$ is:$\\ a = \frac{g (1 - 2\mu)}{3} \\$Therefore, the correct option is Option 3.