The mathematical formulas are necessarily complex; if you want to get this in layman's terms, IMHO you are better off thinking in terms of pictures.
Remember that we are talking about alternating current. This means that the current measured at a point is a continuously varying value. We _identify_ this continuously varying value with a continuously varying mathematical function, the _sine_ function. The benefit here is that a sine function can be described by only a few numbers: amplitude, frequency, and phase angle.
Similarly the current in each of the phases can be identified with suitable sine functions, and described with a very few parameters.
Generally, we accept that there is only a single frequency in the system (60Hz in North America), so don't bother reporting it.
Additionally, we generally report the RMS voltage or current, rather than the _peak_ voltage or current, though mathematically the _amplitude_ of the sine wave is the peak value. This doesn't matter for the present discussion.
If you simply accept the following as truth (it can be shown mathematically, but the proof is not relevant at the moment): given a set of sine functions, all of the _same_ frequency, but of arbitrary amplitude and arbitrary phase relationship, you can _represent_ each of these sine functions as a _vector_. The amplitude of the sine function is represented by the _magnitude_ (length) of the vector. The relative phase angle of the sine function is represented by the _direction_ (angle) of the vector.
In other words, in the world of our simplifying assumptions, we can _represent_ the values that we are dealing with as line segments with length and direction. The great thing here is that it lets us deal with several rather complex problems in a _graphical_ fashion, one that we can see and intuit quite easily.
For example, consider the neutral current question that we are asking here. By the 'kirchhoff current law' the net current flowing into a 'node' must be zero. This means that the current flowing down the neutral wire from the common point must exactly equal the net current flowing into the common point from the loads, so that the _net_ current flowing into the common point is zero. So we have to answer the question "What is the sum of Ia + Ib + Ic?"
The problem, of course, is that Ia, Ib, and Ic all have different phase angles. So we can't simply add the current magnitudes up; there is some sort of canceling going on here.
But if we say that Ia is represented by a vector of length 8, and direction 0 degrees, Ib with length 10 and direction 120 degrees, and Ic with length 15 and direction 240 degrees, and we use the rules of vector addition to add these up, we get that the net current flowing into the neutral from the loads will be represented by a vector of length 6.24 and angle of 224 degrees. The _graphical_ representation of a vector is simply a line segment with length and direction, and the graphical representation of vector addition is simply placing these line segments 'head to tail'.
-Jon
