(a) By conservation of energy
`Q_(1)= W+Q_(2)+Q_(3)`
`i.e. Q_(2)+Q_(3)=Q_(1)-W=1200-200=1000`
And as change in entropy in a reversible-process is zero
`Q_(1)/T_(1)+Q_(2)/T_(2)+Q_(3)/T_(3)=0 i.e.-200/400+Q_(2)/300+Q_(3)/200=0`
i.e., 2Q+3Q=1800
Solving equation (i) and (ii) for `Q_(2)`and `Q_(3)` , we get
`Q_(2)+1200J and Q_(3)=-200J`
i.e. the reservoir at tempertature `T_(2)` absorbs 1200 J of heat while the reservoir at temperature `T_(3)` lose 200 J of heat
(b) Now as change in entropy at constant temperature is given by `DeltaS=(DeltaQ//T)`
So change in entropy of reservoir at temperatures `T_(1),T_(2)` and `T_(3)` will be respectively
`-1200/400=-3J/K,1200/300=4J/Kand (-200)/200=-1J/K`