Find the sum of all 3-digit nos that can be formed by 1, 2

As nothing has been mentioned in the question, we'll assume that repetition of numbers are allowed.
Hence the total number of 3 digit numbers that can be formed from 1,2,3 = 3*3*3
= 27.

Now out of these 27 numbers each of the digits 1,2,3 will occur at each of the hundred's, ten's and unit's position 9 times. e.g. starting from all the numbers having 1 in hundred position we have the following numbers -
111
112
113
121
122
123
131
132
133
Similar would be the sequence for numbers starting with 2 and 3 and if we count we'll find that 1,2,3 occurs at hundreds position 9 times each; at tens position 9 times each and at units position 9 times each.
Hence the sum of all these 27 numbers =
[(1+2+3) * 100 + (1+2+3) * 10 + (1+2+3) * 1 ] * 9
= 6 * 111 * 9 = 54 * 111 = 5994.
Hope this clarifies or else I will elaborate it more.
Thanks!