Find all integer solutions $(x,y,z)\in\mathbb{Z}^3$ that simultaneously satisfy
$$\left\{\begin{array}{l} \displaystyle \frac{1}{3}x+\frac{2}{5}y+\frac{3}{4}z\in\mathbb{Z},\\ \displaystyle \frac{5}{6}x-\frac{3}{7}y-\frac{1}{2}z\in\mathbb{Z}.\\ \end{array}\right.$$
The numbers are only an example. I'm interested in a systemmatic way of solving this kind of problems. The equations are linear and the coefficients are rational. I tried mapping the problem to linear congruential equations using the unit-fraction common divisors
$$\mathrm{gcd}\left(\frac{1}{3},\frac{2}{5},\frac{3}{4},1\right)=\frac{1}{60},\quad \mathrm{gcd}\left(\frac{5}{6},-\frac{3}{7},-\frac{1}{2},1\right)=\frac{1}{42},$$
to obtain
$$\left\{\begin{array}{l} 20x+24y+45z\equiv 0\,(\mathrm{mod}\;60),\\ 35x-18y-21z\equiv 0\,(\mathrm{mod}\;42). \end{array}\right.$$
This problem comes from physics and the solutions $(x,y,z)\in\mathbb{Z}^3$ form a Bravais lattice structure. Physically the ultimate goal is to find the volume of the primitive cell of the lattice. I'm not sure how to proceed. The coefficients of the two congruential equations are not invertible in the $\mathbb{Z}_{60}$ and $\mathbb{Z}_{42}$ rings. Can we do better than enlarging the rings to $\mathbb{Z}_{420}$ and brute-force enumerating all posibilities or are there theories that help?
We have \begin{eqnarray*} \left\{\begin{array}{l} \displaystyle \frac{1}{3}x+\frac{2}{5}y+\frac{3}{4}z\in\mathbb{Z},\\ \displaystyle \frac{5}{6}x-\frac{3}{7}y-\frac{1}{2}z\in\mathbb{Z}.\\ \end{array}\right. \end{eqnarray*} The $x$ and $z$ terms will not be able to create fifths or sevenths so $y \equiv 0 \pmod{35}$ and so the central terms will generate whole numbers. Now consider the first expression, the $x$ term cannot create any quarters and so $z \equiv 0 \pmod{4}$. Finally the second expression no forces $x \equiv 0 \pmod{6}$.
So the general solution is $ \color{blue}{(x,y,z) =(6p,35q,4r)}$ for $ p,q,r \in\mathbb{Z}$.