Answer: $ \large\frac{5}{3}$$mg$

The Free Body diagrams for the problem can be drawn as follows:

Net force in the direction of motion of $m_1$ is $F_1 = T- m_1g =m_1a$ as per Newton's second law.

Net force in the direction of motion of $m_2$ is $F_1 = T- T'=m_2a$ as per Newton's second law.

Net force in the direction of motion of $m_3$ is $F_3 = m_3g- T'=m_3a$ as per Newton's second law.

Solving for acceleration, we get: $a = \large\frac{m_2 + m_3 - m_1}{m_1+m_2+m_3}$$g = \large\frac{2m+3m-m}{m+2m+3m}$$g = \large\frac{2}{3}$$g$

Now, $T-m_1g' = m_1 a \rightarrow T = m_1 (g+a)$.

Subsituting for $a$ and for $m_1 = m$, we get: $T = m (g + \large\frac{2}{3}$$g) = \large\frac{5}{3}$$mg$