Break the code

$\boxed{\color{grey}7} \ \boxed{\color{grey}3} \ \boxed{\color{grey}8}$ -- Clue 4: None is correct ( $\color{grey}\text{grey}$ for an incorrect number).

$\boxed{\color{grey}7} \ \boxed{\color{grey}8} \ \boxed{\color{#3D99F6}0}$ -- Clue 5: One correct number but wrongly placed ( $\color{#3D99F6}\text{blue}$ for a correct number but wrongly placed). Since 7 and 8 are incorrect, 0 must be the correct number but wrongly placed.

$\boxed{\color{#D61F06}6} \ \boxed{\color{grey}8} \ \boxed{\color{#D61F06}2}$ -- Clue 1: One correct number and rightly placed ( $\color{#D61F06}\text{blue}$ for a possible correct number). Since 8 is incorrect, 6 and 2 are possibly correct and rightly placed.

$\boxed{\color{grey}6} \ \boxed{\color{#D61F06}1} \ \boxed{\color{#D61F06}4}$ -- Clue 2: One correct number and wrongly placed. The number 6 cannot be the correct and rightly placed number in clue 1, because 6 remains at the same place in clue 2 and the correct number in clue 2 is wrongly placed. Therefore, 1 and 4 are possibly correct numbers but wrongly place. The clue 1 becomes $\boxed{\color{grey}6} \ \boxed{\color{grey}8} \ \boxed{2}$ ( black for a correct and rightly placed number).

$\boxed{\color{#3D99F6}2} \ \boxed{\color{#3D99F6}0} \ \boxed{\color{grey}6}$ -- Clue 3: Two correct numbers but wrongly placed. Since 6 is incorrect, 2 and 0 must be correct but wrongly placed. Since 2 is on the right, 0 can only be on the left, since the center position is wrong for it. Since the correct number in clue 2 is wrongly placed, that number cannot be 1 but 4. Therefore, the correct answer is $\boxed{A} \ \boxed{B} \ \boxed{C} = \boxed{0} \ \boxed{4} \ \boxed{2}$

$\implies BAC = \boxed{402}$

738 (Nothing is correct). So 7 and 8 are not part of the code.

780 (One number is correct but wrongly placed). 7 and 8 have just been eliminated, so 0 is the number and it is not last.

206 (Two numbers are correct but wrongly placed). 0 is correct, we know that, now we know that it is not in the middle.

The first number of the code is 0.

206 (Two numbers are correct but wrongly placed). One of them is 0, the other could be 6. Let's check that.

682 (One number is correct and well placed) and 614 (One number is correct but wrongly placed). 6 cannot be the number, because the first position cannot be both true and false.

One of the numbers in the code is 2. 682 (One number is correct and well placed). So 2 has to be the last digit of the code.

614 (One number is correct but wrongly placed). Only the central position is to be filled. It has to be 4, since 1 would be correctly placed.

The code is 042.

The answer is $\boxed{402}$

Is this idea copied from a board game called "mastermind".