Show that A(2,-1,-2) , B(1,2,1) , C(2,3,0) , D(5,0,-6) lie…

Show that $A(2,-1,-2)$, $B(1,2,1)$, $C(2,3,0)$, $D(5,0,-6)$ lie in one plane.

Explanation of Method

Four points lying in a common plane.

\[ \overrightarrow{AB}=\{-1,3,3\},\qquad \overrightarrow{AC}=\{0,4,2\},\qquad \overrightarrow{AD}=\{3,1,-4\}. \]

\[ \begin{vmatrix}-1&3&3\\0&4&2\\3&1&-4\end{vmatrix}=-1(-16-2)-3(0-6)+3(0-12)=18+18-36=0. \]

\[[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=0\ \Longrightarrow\ A,B,C,D\ \text{are coplanar}\]

(The tetrahedron $ABCD$ has volume $0$.) One can also check that $\overrightarrow{AD}=-3\,\overrightarrow{AB}+\tfrac52\,\overrightarrow{AC}$.

$A$ $B$ $C$ $D$ $O$

The three vectors from $A$ lie in one plane (shaded), so the four points are coplanar.

\[\boxed{\,[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=0\ \Longrightarrow\ A,B,C,D\ \text{are coplanar}\,}\]