Cho các số thực a, b, c thỏa mãn ${{\left( a-3...

Theo giả thiết: ${{2}^{a}}={{6}^{b}}={{12}^{-c}}\Rightarrow \left\{ \begin{aligned}
& {{2}^{a}}={{12}^{-c}} \\
& {{6}^{b}}={{12}^{-c}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\left( {{2}^{a}} \right)}^{b}}={{\left( {{12}^{-c}} \right)}^{b}} \\
& {{\left( {{6}^{b}} \right)}^{a}}={{\left( {{12}^{-c}} \right)}^{a}} \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& {{2}^{ab}}={{12}^{-bc}} \\
& {{6}^{ab}}={{12}^{-ca}} \\
\end{aligned} \right.\Rightarrow {{12}^{ab}}={{12}^{-bc-ca}}$
$\Rightarrow ab=-bc-ca\Rightarrow ab+bc+ca=0\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{\left( a+b+c \right)}^{2}}={{M}^{2}}$
Do đó, ${{\left( a-3 \right)}^{2}}+{{\left( b-3 \right)}^{2}}+{{\left( c-3 \right)}^{2}}=18\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-6\left( a+b+c \right)+9=0$
$\Leftrightarrow {{M}^{2}}-6M+9=0\Leftrightarrow M=3$, Vậy $M=3$.