Given, $ x \equiv 1(mod3) $

$ \therefore There \ exists \ an \ interger \ l, such\ that \ x = 3l + l ---------1 $

Given, $ x \equiv 2(mod5) $

$ 3l + l $\equiv$ 2(mod5) --------from \ 1$

3l $\equiv$ l (mod 5)

l $\equiv$ 3l (mod 5)

2 $\equiv$ 6l (mod 5 ) ---------------- (multiplying by 2)

2 $\equiv$ (5l + l) (mod 5 )

2 $\equiv$ l (mod 5)

l $\equiv$ 2 (mod 5)

There exists an integer k, such that l = 5k +2 --------2

from 1 and 2

x $\equiv$ 3(sk+2)+1

x $\equiv$ 15k +7 ---------3

also given, x $\equiv$ 3 (mod 7)

$ 15k + 7 \equiv 3 (mod7)-------------from 3 $

$ 15k \equiv -4 (mod7) $

$ 15k \equiv 3 (mod 7)$

$ 3 \equiv 15k (mod7) $

$ 3 \equiv (14k + k) (mod7) $

$ 3 \equiv k (mod7) $

$ k \equiv 3 (mod7) $

There exists an integer m, such that k = 7m + 3 ---------4

from 3 & 4, x = 15 (7m + 3)+7

x = 105m + 52

for m = 0, x= 52

hence, $ x \equiv 52 $ is the solution of $ x \equiv 1 (mod3) $, $ x \equiv 2 (mod5) , x \equiv 3 (mod7) $