Find the smallest positive integer $n$ such that $$\left\{ \begin{array}{l} n\equiv 1\pmod{3} \\ n\equiv 3\pmod{5} \\ n\equiv 5\pmod{7} \end{array} \right.$$
Because $3$, $5$, and $7$ are pair-wise co-prime, we can apply CRT to solve this system. $$\begin{array}{l} 3\times 5\times 7 = 105 \\ (5\times 7)^{-1} \equiv 35^{-1} \equiv 2 \pmod{3} \\ (7\times 3)^{-1}\equiv 21^{-1}\equiv 1\pmod{5} \\ (3\times 5)^{-1} \equiv 15^{-1} \equiv 1\pmod{7} \end{array}$$
Therefore, by CRT, we have the solution as $$n\equiv 1\times 35\times 2 + 3\times 21\times 1+ 5\times 15\times 1\equiv 103 \pmod{105}$$
The smallest positive integer satisfying this relation is $\boxed{103}$.
Meanwhile, this problem can also be solved without employing CRT. We note that all the residues are $2$ less than the modulo. Therefore the answer must be $2$ less the least common divisor of $3$, $5$, and $7$ which is $\boxed{103}$.