A number is multiple of 15 when the number is divisible by 5 and sum of digits of the number is divisible by 3.

Among 1, 2, 3, 4, 5, 6, 7 unit place is filled with 5 so it is multiple of 5.

Now to make it divisible by 3, take remaining 5 digits such a way that sum becomes divisible by 3.

Remaining 5 digits can be

(1) 1, 2, 3, 4, 6

Here sum = 1 + 2 + 3 + 4 + 6 + 5 = 21 (divisible by 3)

This 5 digits can be filled in those 5 placed without repetition in 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 = 51 = 120 ways

(2) 2, 3, 4, 6, 7

Here sum = 2 + 3 + 4 + 6 + 7 + 5 = 27 (divisible by 3)

$$\therefore$$ Number of ways = 51 = 120

(3) 1, 2, 3, 6, 7

Here sum = 1 + 2 + 3 + 6 + 7 + 5 = 24 (divisible by 3)

$$\therefore$$ Number of ways = 51 = 120

$$\therefore$$ Total possible 6 digit numbers divisible by 15

= 120 + 120 + 120 = 360