Then find the sum of all possible values of $abc$. Let $a, b, c$ be positive integers with $0 < a, b, c < 11$. If $a, b, $ and $c$ satisfy
\begin{align*}
3a+b+c&\equiv abc\pmod{11} \\
a+3b+c&\equiv 2abc\pmod{11} \\
a+b+3c&\equiv 4abc\pmod{11} \\
\end{align*}then find the sum of all possible values of $abc$.
What I tried:
I only got this far ...
If the equations from top to bottom were labeled $(1),(2),(3)$,
$(1)+3(2)+(3):$
$7a+11b+7c\equiv 11abc\pmod{11}$
$7(a+c)\equiv 0\pmod{11}$.
Since 7 and 11 are coprime, this means $a+c\equiv0\pmod{11}$ or rather just $a+c=11$ in the given range. Note that because 11 is prime, than modular division by any number not divisible by 11 is allowed henceforth.
Back in $(2)$:
$3b+11\equiv3b\equiv2abc\pmod{11}$
$3\equiv 2ac\pmod{11}$
$2ac=2a(11-a)=22a-2a^2\equiv-2a^2\equiv3\pmod{11}$
$2a^2\equiv-3\equiv8\pmod{11}$
$a^2\equiv 4\pmod{11}$
I just got this far ...

All equations below are understood to be modulo $11$.
Add the $3$ equations together
\begin{eqnarray*}
5(a+b+c)=7abc. 
\end{eqnarray*}
Multiply the first equation by $5$
\begin{eqnarray*}
4a+5b+5c=5abc. 
\end{eqnarray*}
Eliminate $b+c$ and divide by $2a$
\begin{eqnarray*}
10a=9abc.  \\
bc=6.
\end{eqnarray*}
Similarly $ac=7$ and $ab=5$, now multiply these three equations gives $a^2b^2c^2=1$.
Now square root this gives $abc=1$ or $abc=10$.
