Correct answer is (b) 2A, 1A, 0.57A
To explain: The two meshes which contain the 3A source, act as a supper mesh. The mesh equations are:
I1-I3=-3
4I1-14I2+11I3=10
4I1-28I2+10I3=0
Solving these equations simultaneously, we get the three currents as
I1=-1A, I2=0.57A, I3=2A
So currents are 2A, 1A, 0.57A.