We can observe that we have equations:
Solving it we can have B=0 or 5
5 is not taken because, for satisfying the condition A\(\times\)3=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.
we can satisfy this condition only if A=5 i.e., 5\(\times\)3=15.
So we have A=5, B=0 and C=1