What is the largest number n so that n, 2n, and 3n together contain every digit from 1-9 exactly once?

Analytical solution preferred.

A few observations

a) n must be three digits. Two digits is too little and 4 are too many.

b) n must be greater than 123.

c) n must be less than 330. Otherwise, either digits are repeated or 3n has 3 digits.

d) if n, 2n and 3n contain all digits, then their sum of digits = 45, so the sum is divisible by 9.

but the sum = 6n, so n must be divisible by 3.

d) So, that leaves less than 70 that might work