Let $y_A$ and $y_B$ be the positions of blocks A and B.
1. Upper Cord: The length is constant. Thus, if A moves up by distance $x$, B must move down by distance $x$. Their accelerations are equal in magnitude and opposite in direction: $a_A = a$ (up) and $a_B = a$ (down).
2. Lower Cord & Pulley Q: Pulley Q hangs in the loop of the lower cord connecting A and B. The position of Q is the average of the positions of the cord ends (bottoms of A and B). $$y_Q = \frac{y_A + y_B}{2} + \text{constant}$$ Differentiating twice with respect to time: $$a_Q = \frac{a_A + a_B}{2}$$ Since $a_A = -a_B$, we get $a_Q = 0$.
Conclusion: Pulley Q and Block C (attached to it) remain motionless. $a_C = 0$.

Let $T_1$ be the tension in the upper cord and $T_2$ be the tension in the lower cord.

For Block C (4 kg): Since $a_C = 0$, the forces balance. The upward force is $2T_2$ (from the pulley Q). $$2T_2 = m_C g = 40 \, \text{N} \implies T_2 = 20 \, \text{N}$$
For Block A (3 kg): (Assumed accelerating up) $$T_1 – T_2 – m_A g = m_A a$$ $$T_1 – 20 – 30 = 3a \implies T_1 – 50 = 3a \quad \text{— (i)}$$
For Block B (6 kg): (Assumed accelerating down) $$m_B g + T_2 – T_1 = m_B a$$ $$60 + 20 – T_1 = 6a \implies 80 – T_1 = 6a \quad \text{— (ii)}$$

Add equations (i) and (ii): $$(T_1 – 50) + (80 – T_1) = 3a + 6a$$ $$30 = 9a \implies a = \frac{10}{3} \, \text{m/s}^2$$
Now substitute $a$ back into (i) to find $T_1$: $$T_1 = 50 + 3\left(\frac{10}{3}\right) = 60 \, \text{N}$$