Both 'a' and '&a' will point to the same starting address of the array. Moreever '&a[0]' will also point to the same starting address. but because of the pointer/address arthimetic when we increment 'a' and '&a[0]' by 1 it will point to the next element in the same array. Why because p+1 = (int)p + sizeof(*p), and hence the type of p matters for pointer increment.
When we increment '&a' by 1 it will point to the address which is next position after the array. Because here sizeof(*p) = sizeof*&a) = sizeof(a).
So answer should be 70.