First of all, the Nim-sum of 1, 11, and 101 is 111. Not 101.

Secondly, in order to make the explanation avaiable to all people and not only people with knowledge of the binary system:

The trick is to look at each number of dots in a row as a sum of powers of 2. For example 14 = 8 + 4 + 2, and 9 = 8 + 1. Always make the largest power of 2 into a group (so not 9=4+2+2+1 for example).

Look at all groups in all rows and count how many groups of 1 there are, how many of 2, how many of 4, 8, 16, ...

Now for my explenation I will not use the Misere game. The one that erases the last dot wins. The player that can make it so that there are an even number of groups of one size, wins. So in your example:

3 = 2 + 1

7 = 4 + 2 + 1

11 = 8 + 2 + 1

We have one group of 8, one group of 4, three groups of 2 and three groups of 1. We want to have an even number of groups of a certain size. By removing 7 dots from the third row we get:

3 = 2 + 1

7 = 4 + 2 + 1

4 = 4

Now we have two groups of 4, two groups of 2 and two groups of 1. This is what we wanted. The player whose turn it is now will lose. No matter what he does, you can always make it an even number of groups per size. Say the he erases 6 dots form the second row:

3 = 2+1

1 = 1

4 = 4

You remove 2 dots from row three:

3 = 2+1

1 = 1

2 = 2

And now it'll be clear what to do.