I have absolutely no idea what your post means.
In general if $A,~B,~\&~C$ are three non-colinear then the plane determined is:
Let $R=<x,y,z>$ then the plane is $\left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right) \cdot \left[ {\overrightarrow R - \overrightarrow A } \right] = 0$
in order to satisfy the condition r dot n = a dot n . the n which is the normal to the plane must originate from point a am i right? so the point a must be connected DIRECTLY to point B and point P , SO the angle between BAP can be found, but for photo 3, the point A is connected to B , but point P is connected to point B , but not A !. please correct me if my concept is wrong. VECTOR almost making me crazy.