<s>[INST] What does the minimum x that satify $x=24n+12=15m+6=11k+2$? My attempt so far was: let $x=24n+12=15m+6=11k+2$ 
find $x$ as the form $x=24*15*a+24*11*b+15*11*c$
$$2\equiv 24\cdot 15\cdot a\quad (mod\quad 11)\quad \Rightarrow 1\equiv 4\cdot a\Rightarrow \quad a=3\\ 6\equiv 24\cdot 11\cdot b\quad (mod\quad 15)\quad \Rightarrow \quad 2\equiv 8\cdot 11\cdot b\quad (mod\quad 5)\quad \Rightarrow 2\equiv 3\cdot b\quad (mod\quad 5)\quad \quad $$ and I stack here after finding $a$
 [/INST] $\!x\equiv 12\pmod{\!24}\!\iff x/3\,\equiv\ \ \  4\,\pmod{\! 8}$
$\!\!\left.\begin{align} 
&x\equiv \  6\!\!\pmod{\!15}\!\iff x/3\,\equiv\ \ \  2\!\!\!\pmod{ 5}\\ 
&x\equiv \  2\!\!\pmod{\!11}\!\iff x/3\,\equiv -3\!\!\!\pmod{\!\!11}\\
\end{align}\right\}\!\!\!\iff\!\dfrac{x}3\equiv -3\pmod{\!55}\!\iff\! \dfrac{x}3 = \color{#0a0}{-3\!+\!55j}$
$\!\!\!\bmod \color{#c00}8\!:\, 4\equiv \dfrac{x}3\equiv \color{#0a0}{-3\!+\!55}\color{#c00}j\equiv 5\!-\!j\!\!\iff\! \color{#c00}{j\equiv 1}\!\iff\! \dfrac{x}3=-3\!+\!55(\color{#c00}{1\!+\!8i})\!\iff\!\! x\equiv 156\!+\!1320i $
</s>