To make the three-digit 7 □□□□□ divisible by 2, 3 and 5 at the same time, it must be 0 in the unit, because 7+0=7, 7+2=9, 9 is a multiple of 3, 7+5= 12, 12 is a multiple of 3, and 7+8 =/kloc-.

Therefore, its last two digits can be filled with 2, 5 and 8, and the digits can only be filled with 0;

So the answer is: ten digits can be filled with 2, 5, 8, and one digit can only be filled with 0.