Given information:
- A reversible engine absorbs 1200 joules from a reservoir at 400 K.
- The engine performs 200 joules of mechanical work.
- The engine undergoes an integral number of complete cycles.

Assumptions:
- The engine is operating in a Carnot cycle, which consists of two isothermal and two adiabatic processes.
- The engine is in thermal equilibrium with the reservoirs during the isothermal processes.
- The engine is perfectly reversible, meaning there are no losses due to friction or other inefficiencies.

Explanation:
To find the sum of quantities of heat exchanged with the other two reservoirs, we need to consider the energy conservation in the reversible engine.

1. First Law of Thermodynamics:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

2. Carnot Cycle:
In a Carnot cycle, the engine operates between two heat reservoirs at different temperatures. The efficiency of a Carnot engine is given by:

η = 1 - (T2 / T1)

where η is the efficiency, T1 is the temperature of the hot reservoir, and T2 is the temperature of the cold reservoir.

3. Applying the First Law:
In this problem, the engine absorbs 1200 joules of heat from the hot reservoir and performs 200 joules of work. Therefore, the change in internal energy of the engine is:

ΔU = 1200 J - 200 J = 1000 J

4. Efficiency of the Engine:
The efficiency of a reversible engine is given by:

η = W / Qh = 200 J / 1200 J = 1/6

5. Applying Carnot Efficiency:
Since the engine is operating in a Carnot cycle, its efficiency is also given by:

η = 1 - (T2 / T1)

Let's assume that T2 is the temperature of the cold reservoir. Rearranging the equation, we get:

T2 = T1 - T1 * η

Plugging in the given values, we have:

T2 = 400 K - 400 K * (1/6) = 400 K - 200/6 K = 400 K - 33.33 K = 366.67 K

So, the temperature of the cold reservoir is 366.67 K.

6. Heat Exchange with the Cold Reservoir:
During the isothermal process with the cold reservoir, the engine releases heat to the reservoir. The amount of heat released can be calculated using the equation:

Qc = W / η = 200 J / (1/6) = 1200 J

Therefore, the engine releases 1200 joules of heat to the cold reservoir.

7. Heat Exchange with the Hot Reservoir:
Since the engine is reversible, the heat exchange with the hot reservoir can be calculated using the equation:

Qh = Qc + W = 1200 J +