How to efficiently use a calculator in a linear algebra exam, if allowed

There is a very nice trick for showing that such matrix has full rank, it can be performed in a few seconds without any calculator or worrying "moral bending". The entries of $M$ are integers, so the determinant of $M$ is an integer, and $\det M\mod{2} = \det(M\mod{2})$. Since $M\pmod{2}$ has the following structure $$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} $$ it is trivial that $\det M$ is an odd integer. In particular, $\det M\neq 0$ and $\text{rank}(M)=3$.

You're allowed to use your calculator. So, if it were me on the test, I'd write something like this:

$$ \pmatrix{5&6&7\\12&4&9\\1&7&4} \overset{REF}{\to} \pmatrix{ 1 & 0,3333 & 0,75\\ 0 &1 &0,75 \\ 0 & 0 & 1 } $$ because the reduced form of $M$ has no zero rows, $M$ has rank $3$.

REF here stands for row-echelon form.

Note: You should check with your professor whether or not this constitutes a sufficient answer. It may be the case that your professor wants any matrix-calculations to be done by hand. See Robert Israel's comment.

If that's the case, then you should do row-reduction by hand. It only takes 3 row operations, though.