$6$ preference order for voter are possibe:
$ABC.ACB,BCA,BAC,CAB,CBA$ also Given that
$a=ABC+ACB+CAB(A$ prefer over $B) ---(1)$
$b=BCA+BAC+ABC(B$ prefer over $C) ---(2)$
$c=CAB+CBA+BCA(C$ prefer over $A) ---(3)$
Adding $1,2$ and $3$ we get
$a+b+c=2(ABC+BCA+CAB)+ACB+BAC+CBA$
Now we know that $ABC+ACB+BAC+BCA+CAB+CBA=1$ therefore
$[ABC+ACB+BAC+BCA+CAB+CBA]<[2(ABC+BCA+CAB)+ACB+BAC+CBA]<2(ABC+ACB+BAC+BCA+CAB+CBA)$
Hence we can say that value of $a+b+c$ must be between $1$ and $2$
option (c) value greater than $2$ hence correct answer is (c)