If Q3 is to remain in place then the net force on that charge will have to be 0. If you think about this, since forces are vectors, you cannot have Q3 as the third point of a triangle because there will always be a component of the net force directed either toward or away from the line containing the other two charges.

No, Q3 will have to lie somewhere on the same line as Q1 and Q2. Since we don't even know what the sign of the charge of Q3 will be (though logic says it must be negative) I will simply call the position of Q3 as x. (The y coordinate is 0, of course.) I am going to assume for the moment that Q3 is between charges 1 and 2 until I run into a contradiction. So our assumptions are the Q3 is negative and between the other two charges.

So let's calculate the magnitude of the force on Q3 by Q1:

Since this force is toward Q1 the force component is negative:

The magnitude of the force on Q3 by Q2 is:

Since this force is toward Q2 the force component is positive:

Now the sum of these forces is required to be 0:

We may solve this equation for x:

<-- Cancel out the extra stuff:

Typically Physics equations don't factor, but this one does:

So x = 3 cm or x = -9 cm.

Does the x = -9 cm solution make any sense? No, because at this point both Q1 and Q2 would be pulling it in the +x direction. So we discard this solution.

So let's look at the net force on Q1 (we could also look at Q2):

This force is repulsive (pushing Q1 in the -x direction) so the force component is negative:

and

This force is attractive (pulling Q1 in the +x direction) so the force component is positive:

And again the net force on Q1 is 0:

We may solve this equation for Q3/q:

I'll let you verify that the net force on Q2 is also 0.

-Dan