The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $p\mid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
- If $p\nmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}\equiv b^{p-1}\equiv c^{p-1}\equiv 1 \pmod{p}$. Thus, $p\mid 3$, hence $p=3$.
- If $p\mid a$, and $p\nmid b,c$, then we obtain $p\mid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+c\mid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+c\mid 3abc$. Next, note that, if $9\mid a+b+c$, then $3\mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3\mid a$, then using $a+b+c\mid 2(ab+bc+ca)$, we obtain that $3\mid bc$, hence $3\mid b$ or $3\mid c$. However, this, together with $3\mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9\nmid a+b+c$.
Similarly, if $4\mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4\mid a$. But this gives, $4\mid 2(ab+bc+ca)$, yielding $2\mid (ab+bc+ca)$, yielding $2\mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $p\mid 3abc$ implies $p\mid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $p\mid ab+bc+ca$, together with $p\mid ab+ac$ implies $p\mid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9\nmid a+b+c$, and finish.
A quick computer check shows that $7272$, $8484$ and $9696$ have the same (maximum) number of divisors ($24$):
{1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72, 101, 202, 303, 404, 606,
808, 909, 1212, 1818, 2424, 3636, 7272},
{1, 2, 3, 4, 6, 7, 12, 14,
21, 28, 42, 84, 101, 202, 303, 404, 606, 707, 1212, 1414, 2121,
2828, 4242, 8484},
{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96, 101,
202, 303, 404, 606, 808, 1212, 1616, 2424, 3232, 4848, 9696}
so if I understand your question (which of these is minimum?), the answer is: $7272$.
The original question did not state that the solution must be done by hand, but in that case, @MishaLavrov's approach (below) is the best:
Recognize that $abab = ab \cdot 101$, so the factors of $abab$ are maximized when $ab$ has the most factors. It is a simple matter to find (by hand) that $72$, $84$ and $96$ have the same (maximum) number of factors.
The rest is easy.
Best Answer
Assuming Zunayed has more chocolates than Pial and at the end it is not required to have $P<Z<B$, but only the intermediate steps. (Pial is not to blame if Zunayed decided to give away chocolates to Bindu).
Can Zunayed have 3 chocolates? No, because Pial can give 9 chocolates so that $Z=12>11=P$. (According to the condition, Pial must give 10 and keep $\color{red}{10}$).
Can Zunayed have 2 chocolates? Yes, Zunayed has $\color{red}{2}$ chocolates.
Can Zunayed have 1 chocolate? No, because Zunayed gets 10 from Pial to have 11 in total, however, Zunayed can not give half of it to Bindu (it is assumed a chocolate is indivisible).
First note that Zunayed has 12 chocolates after receiving 10 from Pial.
Can Bindu have 3 chocolates? No, because Zunayed has 12 chocolates and Zunayed can give 5 to Bindu, because $B=8>7=Z$. (According to the condition, Zunayed must give 6).
Can Bindu have 2 or 1 chocolates? Yes, Bindu has $\color{red}{2}$ or $\color{red}{1}$ chocolates.
Hence, the total number of chocolates is $\color{red}{24}$ or $\color{red}{23}$. That is: $$P=20,Z=2,B=2 \ \ \text{or} \ \ P=20,Z=2,B=1$$ in the beginning.