Hint $\ $ Note that by the constant-case optimization of CRT we have
$$\rm\begin{eqnarray} x\equiv 2\pmod 4\\\rm x\equiv 2\pmod 5\end{eqnarray}\ \iff\ x\equiv 2\pmod{4\cdot 5}$$
Now applying Easy CRT to the RHS congruence and the remaining congruence we obtain
$$\rm\begin{eqnarray} x\equiv \color{#C00}2\pmod{\color{blue}{20}}\\\rm x\equiv \color{#0A0}{12}\pmod{27}\end{eqnarray}\ \iff\ x\equiv \color{#0A0}{12}+27\,\left[\dfrac{\color{#C00}2\!-\!\color{#0A0}{12} }{27}\ mod\ \color{blue}{20}\right]\equiv 282\pmod{\color{blue}{20}\cdot 27}$$
where the above fraction is computed as $\rm\, mod\ 20\!:\ \dfrac{10}{27}\equiv \dfrac{10}{7}\equiv\dfrac{30}{21}\equiv \dfrac{10}1,\ $ and $\rm\:-10\equiv 10$
The CRT basically says says that the natural projection from $\Bbb Z$ onto that product is surjective.
Being surjective, $z\mapsto(0+5\Bbb Z,0+11\Bbb Z,1+7\Bbb Z)$ for some $z\in \Bbb Z$. By an isomorphism theorem, you turn this into a isomorphism from $\Bbb Z/(385)$ to the product. It is still surjective, and since it is injective, this says the answer $z$ is unique (mod $385$).
One naive way to solve this would be to look at multiples of $55$ and see which one is congruent to $1$ mod $7$.
There is a simple algorithm to compute more complex cases. Let's suppose you want a simultaneous solution to get $(a,b,c)$ instead of $(0,0,1)$. Since 77 is coprime to 5, you can find an inverse mod 5, which turns out to be 3. Then $77\cdot3\cdot a\cong a\pmod{5}$, but it is zero mod 7 and mod 11.
Since 35 is coprime with 11, we can find an inverse mod 7, which turns out to be 6.Then $35\cdot 6\cdot b\cong b\pmod{11}$, but it is zero mod 5 and mod 7.
Finally, 55 has an inverse mod 7, namely 6. Thus $6\cdot 55\cdot c\cong c\pmod{7}$, and zero mod 5 and mod 11.
Summing all of these, we get that $77\cdot 3\cdot a+35\cdot 6\cdot b+55\cdot 6\cdot c=z$ is a solution to the simultaneous modular equations. If you need it to be below 385 you can just reduce it mod 385, or if you need it to be some number you can just increase it mod 385 to get the desired number.
Best Answer
The Chinese remainder theorem tells you that given a system of linear congruence of the form $$x\equiv a_k\pmod{m_k}\ k=1,2,\dots,N$$ where the moduli $m_1,m_2,\dots,m_N$ are pairwise relatively prime, the system has exactly one solution modulo $m_1m_2\cdot\dots\cdot m_N$. You have such a system, and you can see that the proposed solution satisfies each of the congruences, so it is the only solution modulo the product of the moduli, which is $n$ in your case.