Given $0<a<b<c$ three distinct integers, prove that there exists infinity many numbers $n$ such that $a+n,b+n,c+n$ are relatively prime to each other.
For the case when $(a,b)=(b,c)=(a,c)=1$ which means they are already relatively prime the answer is easy, i found that $n= (c-b)(c-a)(b-a)k $ for $k \in \mathbb{N} $ works, for instance $gcd((c-b)(c-a)(b-a)k + a, (c-b)(c-a)(b-a)k +b) = (b-a,(c-b)(c-a)(b-a)k +a)= (b-a,a) = (b,a) = 1$
And so for the rest.
But what to do when one or more are not relatively prime ?
How to solve it in more general way?
Note: this does not work with $4$ numbers. For example, if you start with $(2,3,4,5)$ then whatever you add to each term, two of the set will be even. Whatever argument one constructs must fail if you try to extend it to $4$ terms.
Your remark shows that it would suffice to find a single such $n$.
Without loss of generality, suppose that $a>b>c$.
Note: if a prime $p$ divides $\gcd (a+n,b+n)$ then $p\,|\,a-b$.
Use the Chinese Remainder Theorem to solve the system of congruences $n\equiv 1-a\pmod {p_i}$ for each of the primes $p_i$ which divide $a-b$ or $a-c$, and $n\equiv 1-c\pmod {q_j}$ for each of the primes $q_j$ which divide $b-c$. Note that if $p_i=q_j$ for some pair $(i,j)$ then the congruences are consistent. (N.B. This is the bit that fails if you extend to $4$ terms).
We claim that $n$ works.
Indeed, If a prime $\psi$ divides $\gcd(a+n,b+n)$ then $\psi\,|\,a-b$ but in that case $a+n\equiv 1 \pmod {\psi}$, a contradiction.