How to solve the pair of equations $4a-11b+12c=22$ and $a+5b-4c=17$ over the integers?

Given $7a+4b=73$ you can take it $\bmod 7$ to get $4b \equiv 3 \bmod 7, b \equiv 6 \bmod 7$. This gives $b=6$ or $13$ because $20$ is too large. Plugging in, you find $13$ is too large as well.

If you want to do it without the modulo operation, note that $a \lt 11$ because otherwise $7a \gt 73$, so you only have ten choices and can try them all.

Once you find one solution in integers, any other solution happens by adding an integer multiple of the coefficient cross product, namely $$ \langle -16, 28, 31 \rangle $$ You know the solution $ \langle 7,6,5 \rangle \; . $ Any other integer solution is $$ \langle 7-16t, 6+28t, 5+31t \rangle \; . $$ If $t > 0$ we get $7 - 16 t < 0.$ If $t < 0$ then $6 + 28 t < 0.$ It follows that $t=0,$ the only solution in positive integers is the given one. There are infinitely many integer solutions, they lie on the line I described, but that line passes only briefly through the first (positive) octant.