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.
Given a set of primes, $\{p_i \mid i\in[1,k]\}$, compute a basis: A set of integers $b_i$ such that $b_i \cong 1 \mod p_i$ and $b_i \cong 0 \mod p_j$ for $j \neq i$. Since they're primes, they are coprime, so there is a linear combination of them that produces one, and this is what you need to find your basis.
Suppose then that some number, $n$, is simultaneously congruent to $a_i \mod p_i$ for the set of primes. Then $n \cong \left( \sum a_i b_i \right) \mod \prod_{i=1}^k p_i$.
(Watching how all that happens, the basis does act as a matrix to multiply the vector of the various residues.)
Observation: The set of primes must be large enough and/or contain enough primes that the resulting product is smaller than the product of the primes. Otherwise the results "roll over" and you will lose uniqueness (unless you have some additional way to keep track of how many times the result has rolled over. An easy way to do that is add a bigger prime).
Best Answer
If $p\equiv 1 \bmod 4$ because $\Bbb{Z}_p^\times$ is cyclic with $p-1$ elements there is $c^2\equiv -1 \bmod p$ so that
$$\mathbb{Z}[i]/(p)\cong \mathbb{Z}[x]/(x^2+1,p)\cong \mathbb{Z}_p[x]/(x^2+1) \cong \mathbb{Z}_p[x]/(x+c)(x-c)$$ $$\cong \mathbb{Z}_p[x]/(x+c)\times \mathbb{Z}_p[x]/(x-c)\cong \mathbb{Z}_p\times \mathbb{Z}_p$$ where the middle isomorphism is based on $\frac{p+1}{2}c (x-c)\equiv 1 \bmod (x+c),\equiv 0 \bmod (x-c)$