## Let $f(n)$ be the number of $6$-tuples $(x_1,x_2,x_3,x_4,x_5,x_6)$ such that:

All $x_i$ are integers with $0 \leq x_i < n$
$\gcd(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2,\ n^2)=1$
Let $\displaystyle G(n)=\displaystyle\sum_{k=1}^n \frac{f(k)}{k^2\varphi(k)}$
where $\varphi(n)$ is Euler's totient function.
For example, $G(10)=3053$ and $G(10^5) \equiv 157612967 \pmod{1\,000\,000\,007}$.
Find $G(10^{12})\bmod 1\,000\,000\,007$.

### This is a quite complicated problem, and it involves some relatively advanced concepts in number theory. Unfortunately, it is too complex to be fully solved here. But I can start explaining the basic ideas behind the solution.

To count the number of 6-tuples, we’ll use the principle of inclusion-exclusion. This principle is a general formula in combinatorics that calculates the number of elements in a finite union of sets.

In simpler words, it could be formulated as follows: to find out how many elements are there in at least one of the predefined sets, we should first count the total size of the sets, then subtract the sizes of their pairwise intersections, then add back sizes of the intersections of every three, etc.

Here is a sketch of the solution:

1. There are $n^6$ total 6-tuples $(x_1,x_2,x_3,x_4,x_5,x_6)$ where each $x_i$ can take a value from $0$ to $n-1$.

2. We would need to compute the number of 6-tuples such that $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2$ gives a number that has common factors greater than 1 with $n$.

3. Subtract the value from step 2 to $n^6$.

4. Apply modular arithmetic in your computations to avoid results going beyond capacity. Remember that $a \equiv b \mod p$ means that when a and b are divided by p, the remainders are the same. Also dot forget that the product of two residues mod p is equal to the residue of the product of the two numbers.

Once you’ve done these calculations, you will have a good start at computing $G(n)$. This is an outline for a solution, and it requires a lot more details to get to a final answer.

This specific problem feels more like a contest problem and requires more familiarity with number theory to be solved entirely. The Euler’s Totient function and handling the modulus operations in this context can get quite overwhelming even for college-level mathematicians.

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