Here it is given that is divisible by 3.

We know that if a number is divisible by 3, then sum of digits must be a multiple of 3.

∴ 3 + 5+a+6+4 = multiple of 3

∴ a + 18 = 0,3,6,9,12,15,…..

But ‘a’ is a digit, hence, ‘a’ can have value between 0 to 9.

∴ a + 18 = 18 which gives a = 0.

∴ a + 18 = 21 which gives a = 3.

∴ a + 18 = 24 which gives a = 6.

∴ a + 18 = 27 which gives a = 9.

∴ a = 0,3,6,9