= Solution
Using the common denominator $70$, the three given generators become
$$
\frac25=\frac{28}{70},\qquad
\frac37=\frac{30}{70},\qquad
\frac12=\frac{35}{70}.
$$
Thus their <submodule> is contained in $(1/70)\mathbb Z$. For the reverse inclusion, the following explicit <integer> linear combination supplies the missing generator:
$$
2\cdot28-3\cdot30+35=1,
$$
so
$$
2\cdot\frac25-3\cdot\frac37+\frac12=\frac1{70}.
$$
Therefore $1/70$ belongs to the generated <submodule>, and all its <integer> multiples belong as well. We have established equality in both directions:
$$
M=\frac1{70}\mathbb Z.
$$
Finally, $k/70=0$ for an <integer> $k$ forces $k=0$, so the single generator is linearly independent. Hence the required <basis of a module> is
$$
\boxed{\left\{\frac1{70}\right\}.}
$$
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