Suppose that the six digits that are used are a, b, c, d, e and f and we use them to make the three 2-digit numbers : ‘ad’, ‘be’ and ‘cf’. The sum of these numbers is(10a + d) + (10b + e) + (10c + f ) , 

that is10(a + b + c) + (d + e + f ). To make this as large as possible, we need to choose a, b and c so that a + b + c is as large as possible, and then d, e and f  from the remaining digits so that   

d + e + f  is as large as possible.

Clearly, this is achieved by choosing a, b, c to be 7, 8, 9, in some order, and then d, e, f  to be 4, 5, 6  in some order.

So the largest possible sum is 10(7 + 8 + 9) + (4 + 5 + 6) = 255 .