The CRT with three moduli

First two. From the first two: $x\equiv7\pmod{15}$ (verify: $7\equiv1\pmod3$, $7\equiv2\pmod5$).

Combine with third. $x=7+15k$, need $7+15k\equiv3\pmod7$. $15\equiv1\pmod7$, $7\equiv0\pmod7$, so $15k\equiv3\pmod7$, i.e. $k\equiv3\pmod7$.

Solution. $x=7+15(3+7m)=52+105m$.

Conclusion. $x\equiv52\pmod{105}$.

Check. $52\equiv1\pmod3$, $52\equiv2\pmod5$, $52\equiv3\pmod7$. Correct.