Thank you for all the feedback.

So I understand intuitively that the three conductors passing through the GFCI CT are instantaneously summed. This should equal 0 for normal operations. What I am having trouble with is the angle of the neutral.

When adding the three vectors as described above the sum would not be 0.

Correct. We follow Kirchhoff's Current Law. One way to say it is:

The sum of all currents entering a node equals zero (or equivalently, the sum of all currents leaving a node equals zero).

For normal conditions, the super-node only has A, B, C, & N lines crossing the boundary (i.e. 4 possible current paths). For a ground fault, we also have a "G" line (an additional current path).

So for normal conditions, the neutral current into the node at any given instant is the sum of all other currents leaving the node or what I gave in my previous example. To get the neutral current leaving the node, we would indeed add 180 deg to the prior result, as you surmised (210 deg and 270 deg).

For a ground fault, we sum A, B, C, & N to get the current for "G". If G <> zero (or some other threshold), then we trip for a ground fault.