Replace $A,B,C$ by suitable numerals

Hint: Whenever we have this type of problem, try to simplify by considering numbers from unit place. That is in unit place we have $6 + A = 3$, now try to analyze the values on simplifying the $6 + A = 3$ we get $A = 3 - 6 = - 3$ which is not possible. If $A$ is greater than ten then one will be carried over to the tens place hence we can write as $6 + A = 13$ . Therefore we get $A = 13 - 6 = 7$. In the same way, solve for other alphabets.

Complete Step by Step Solution:
Here in this type of problems, try to simplify by considering numbers from the unit place of the given problem. That is in unit place we have $6 + A = 3$, now try to analyze the values, on simplifying the $6 + A = 3$ we get $A = 3 - 6 = - 3$ which is not possible. If $A$ is greater than ten then $1$ will be carried over to the tens place. Hence we can write as $6 + A = 13$. Therefore we get $A = 13 - 6 = 7$. In the same way we solve for other alphabets.
We have a carryover from the unit place so we have at tens place as $1 + B + 9 = 7$. If we simplify this expression we get $B = 7 - 9 - 1 = - 3$ . Again which is not possible, so as we did for $A$, we can write for $B$ as well. Therefore we have $1 + B + 9 = 17$ where $1$ is carried over to the hundredth place. On simplifying we have $B = 17 - 9 - 1 = 7$ .
Now in the hundredth place we have $1 + C + 6 = 1$ , where the left hand side $1$ is carried from tens place. If we simplify this we get $C = 1 - 6 - 1$ which is again not possible. Hence we can write $1 + C + 6 = 11$ and we simplify to get $C = 11 - 6 - 1 = 4$ and one is carried to a thousand places.

Hence we have the value $A = 7$ , $B = 7$ and $C = 4$.

In order to cross verify, we can do as below.

Note:
Whenever we have this type of problem, first we need to start comparing units and be careful regarding carryover values, which play an important role in this type of problems. If you analyze the question properly, you can arrive at the required answer.