I must discover if these points are on the same plane: A=(0;1;-2) B=(-1;2;0) C=(-3;4;-2) D=(-2;3;-2) I know that this implies: $\overrightarrow{AB}$=$\lambda\overrightarrow{AC}$+$\tau\overrightarrow{AD}$
$\left(\begin{array}\\{-1}\\{1}\\{2}\end{array}\right)$=$\lambda\left(\begin{array}\\{-3}\\{3}\\{0}\end{array}\right)$+$\tau\left(\begin{array}\\{-2}\\{2}\\{0}\end{array}\right)$
But the problem is that the z will never be 2 no matter what $\lambda$ or $\tau$. If I switch $\overrightarrow{AC}$ with $\overrightarrow{AB}$ it results, why?
A plane is determined by three points. The plane determined by $A$, $B$, and $C$ is $$P(s,t) = A + s\vec{AB}+t\vec{AC}$$ Notice that $\vec{AB}$ is not a scalar multiple of $\vec{AC}$ so this is in fact a plane (a $2$-dimensional affine space). It is clear that plugging each of $A$, $B$, and $C$ into this equation you can find a solution $(s,t)$ that solves it. But if it's less than clear just notice that $$A=A+0\ \vec{AB}+0\ \vec{AC}\quad \implies A=P(0,0) \\ B=A+1\ \vec{AB}+0\ \vec{AC}\quad \implies B=P(1,0) \\ C=A+0\ \vec{AB} + 1\ \vec{AC}\quad \implies C=P(0,1)$$
Now check whether there's a solution when we plug $D$ in. If there is then $D$ is in the same plane as $A$, $B$, and $C$. If not, then it's not.