How do I check if I found all generators of the solution module of a linear homogeneous diophantine equation?

Question

I have a linear diophtantine equation $$28x + 30y + 31z=365$$ and want to find all its solutions. Using the Gauss-Euclid algorithm, I start with the matrix $$\pmatrix{28 & 1 & 0 & 0 \\ 30 & 0 & 1 & 0 \\ 31 & 0 & 0 & 1}$$ and reduce it to $$\pmatrix{0 & 15 & -14 & 0 \\ 0 & -1 & 3 & -2 \\ 1 & 0 & -1 & 1}$$ which shows me that $\gcd(28, 30, 31)=1$, $31-30=1$ is a linear decomposition of $\gcd$, and $(15, -14, 0), (-1, 3, -2)$ are two linearly independent solutions of the homogeneous system $28x + 30y + 31z=0$. Now, since the equation is linear, solutions of the inhomogeneous system differ by solutions of the corresponding homogeneous system, and all I need to prove is that the two elements I have found generate the module of solutions of the homogeneous equation. If the equation had only two variables, and had the form $$Ax+By=k\cdot \gcd(A, B)$$ then all the solutions of the homogeneous system would differ by $l(\alpha, -\beta)$ where $\alpha A = \beta B = \text{lcm}(A, B)$, $k, l\in \mathbb{Z}$. However, for three variables I have at least two options to obtain a zero linear combination: multiplying by factors of $\text{lcm}(A, B)$ and by factors of $\text{lcm}(B, C)$, i.e. if $A\alpha = B\beta = \text{lcm}(A, B)$, then $A\alpha - B\beta + C\cdot 0 =0$, and an analogous triple of coefficients $(0, \beta', -\gamma')$ from $\text{lcm}(B,C)=B\beta' = C\gamma'$, with the coefficient before $A$ being 0. Now, how do I prove there is no triple $(\alpha'', \beta'', \gamma'')$ such that $$A\alpha'' + B\beta'' + C\gamma'' = 0 $$ and such that for no $k, l\in \mathbb{Z}$ $$k(\alpha, \beta, \gamma)+l(\alpha', \beta', \gamma')=(\alpha'', \beta'', \gamma'')$$? Or, if there is indeed such triple, how do I find it? I checked this and have also checked Bill Debuque's answer to this one, but neither contains a discussion of my question. I guess, it should have something to do with the fact that $\text{lcm}(a, b)$ divides every common multiplier of $a$ and $b$, so $\text{lcm}(a, b)|\text{lcm}(a, b, c)$ and $\text{lcm}(b, c)|\text{lcm}(a, b, c)$, but I don't see how to extract a relevant statement about linear combinations from this fact.