I'm doing this exercise:
Let $3$ different points of an affine plane, with barycentric coordinates $X=(x_0,x_1,x_2), Y=(y_0,y_1,y_2), Z=(z_0,z_1,z_2)$ respect to a fixed reference frame. Prove that those points are coplanar if and only if $$\begin{vmatrix} x_0 & y_0 &z_0 \\ x_1 & y_1 & z_1\\ x_2 & y_2 & z_2 \end{vmatrix}=0.$$
I can't use affine transformations, nor thales or ceva.
I tried to solve the problem by creating a reference frame of three points $A, B$ and $C$, with cartesian coordinates $A=(a_1,a_2)$, $B=(b_1,b_2)$ and $C=(c_1,c_2)$, and then doing the matrices in order to change the system of reference to the canonical frame, but it's very long and I don't know if I'm taking the right way. And in the end I obtain:
$$\begin{vmatrix} x_{1}^{'} & x_{2}^{'} &x_{3}^{'} \\ y_{1}^{'} & y_{2}^{'} & y_{3}^{'}\\ 1 & 1 & 1 \end{vmatrix}=0 \iff \begin{vmatrix} x_0 & y_0 &z_0 \\ x_1 & y_1 & z_1\\ x_2 & y_2 & z_2 \end{vmatrix}=0,$$
where $X=(x_{1}^{'},y_{1}^{'}), Y=(x_{2}^{'},y_{2}^{'}), Z=(x_{3}^{'},y_{3}^{'})$ are the coordinates of $X, Y, Z$ in the canonical frame.
Any hint or indication will be very well recieved. Thanks.
$\textbf{Proof}$: ($\Longrightarrow$) Assume that $\det(X Y Z)=0$. Then, the equation $aX+bY+cZ=0$ has a non-trivial solution. So we can write $Z = \alpha X + \beta Y$ for some constants $\alpha$ and $\beta$ (not both zero).
But $Z$ $\sim$ $Z'$ where $Z'$ is on the line connecting $X$ and $Y$ i.e. $Z' = (1-\lambda)X+ \lambda Y$ for some $\lambda$ i.e. $X$, $Y$, and $Z$ are collinear.
($\Longleftarrow$) Let $X$,$Y$, and $Z$ be collinear. Then, $Z = (1-t)X+tY$ for some $t$. Thus, $\{X,Y,Z\}$ is a linearly dependent set, implying that $det(XYZ)=0$