In the list above, k, m, and n are three distinct positive i

Option E.
Sum of 6 nos.=8*6=48
So k+m+n=48-27=21
Also,median in case of even no. of terms=average of mid two terms=>6.5*2=13=third+fourth terms.
Actually,when we take any of k,m,n to be 5,we get the other two values to be equal which is not possible since
the question specifically asks for DISTINCT integers.

I would use the method of elimination of options here since the question says "cannot be the value". All I have to do is prove that the rest of the options can be values of k, m and n.

Since median is 6.5, the first options I will try are 6 and 7 - two numbers will be eliminated if I can find a case where this works.

3, 4, 6, 7, 20 - the avg of these numbers is 8. If the last number is also 8, the mean will remain 8 as desired. So in fact, we eliminated 3 options here. k, m and n can be 6, 7 and 8.

Let's try 5 now, not 9 because 9 is more complicated. 9 gives us two number less than 6.5 and 2 more than 6.5. So there will be many different options. If instead 5 is in the list, we now have 3 numbers less than 6.5, so the other 3 numbers must be greater than 6.5 and the average of one of those numbers with 5 must be 6.5. So the fourth number should be 8 on the list to give the median 6.5. These 5 numbers (3, 4, 5, 8, 20) give an average of 8. The sixth number must be 8 to keep the average 8 but numbers must be distinct. So this is not possible. Hence none of k, m and n can be 5.

Answer (E)