I have absolutely no idea what your post means.

In general if $A,~B,~\&~C$ arethen the plane determined is:three non-colinear

Let $R=<x,y,z>$ then the plane is $\left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right) \cdot \left[ {\overrightarrow R - \overrightarrow A } \right] = 0$