a+13c+2b+1a+3c+12b=1......(i)
a+11+2b+11+3c+11=2......(ii)
Add (i) and (ii)
⇒a+13c+1+2b+1a+1+3c+12b+1=3
⇒ A.M. ≥ G.M. ⇒(3a+13c+1+2b+1a+1+3c+12b+1)≥(1)31⇒a+13c+1=1 and 3c+12b+1=12b+1a+1=1⇒(a+1)=(2b+1)=(3(+1))=k⇒k1+k1+k1=2⇒k3=2⇒k=23⇒a=21,b=41,c=61⇒a1+b1+c1=2+4+6=12