Question: Balance the following oxidation-reduction reaction in basic condition:
H2O +(S2-)+(MnO4-)--> S+MnS+(OH-)

My answer is:
6H2O +5(S2-)+2(MnO4-)--> 5S+2MnS+12(OH-)
However, thats not an option.

The equation is not balanced , amount of Oxygen and sulfur is not correct.
Reduction

MnO4- + 4 H2O + 5  e- => Mn2+ +  8 OH-

Oxidation

S2- => S + 2  e-

Multiply each equation so the electrons equal

2 MnO4- + 8 H2O + 10  e- => 2 Mn2+ +  16 OH-

5 S2- => 5 S + 10 e-

Addition
2 MnO4-  + 8 H2O + 5 S2- => 2 Mn2+ + 16 OH- + 5 S

Precipitation of MnS need two more S2-

2 MnO4- + 8 H2O + 7 S2- => 2 MnS + 16 OH- + 5 S

Great explanation !