hence

...(1)

from condition ii)

...(2)

now from (1) and (2)

...(*)

ok now take (mod 2)

that means a,b both even or both odd since their sum is even

now (*) mod 3

that means a+b+c is a multiple of 3 so N is a multiple of 3

now take (*) (mod 11)

if you take (mod 4)

you will have

a and b are distinct numbers so the difference between a and b is a multiple of 4,

mod 5

but 3 is primitive with 5 not common divisors between them

so

let's try some numbers let a be "1", b have two choices 5 or 9 false in the each probability

if a "2" , b = 6 , c=3 these fail in equation (mod 3)

if a "3" , b =7 , c =5 fail in (mod 11)

if a "4" , b =8 , c =7 fail in (mod 3)

if a "5" , b =9 , c =4 works in (mod 2,3,4,5,11) now justify the answer

is this true

594 = 33(5)+33(9)+33(4)

33*18 = 594 true

nice question