If D is a point of the plane ABC (which is noncollinear) and ABD is noncollinear, Then prove that the plane ABD is the plane ABC.
If each of a, b, & c is a vector the we know that $\displaystyle a \cdot \left( {b \times c} \right) = \left( {a \times b} \right) \cdot c$.
We also know that given four points A, B, C, D these points co-planar if and only if $\displaystyle \overrightarrow {AD} \cdot \left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right) = 0$
So for any point X on the plane ABC we must have $\displaystyle \overrightarrow {AX} \cdot \left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right) = 0$.
Can you use these to show that the two sets (planes) are equal?