(Too long for a comment.)
You can prove the identity $$\begin{vmatrix} x & y & z & 1\\x_1 & y_1 & z_1 & 1\\x_2 & y_2 & z_2 & 1\\x_3 & y_3 & z_3 & 1 \end{vmatrix}=\det\left(\begin{pmatrix}x & y & z\\x & y & z\\x & y & z\end{pmatrix}-\begin{pmatrix}x_1 & y_1 & z_1 \\x_2 & y_2 & z_2 \\x_3 & y_3 & z_3\end{pmatrix}\right)$$ If the three points given are collinear, then any row you pick can be written as a linear combination of the two other rows, making the determinant zero, and thus you wind up with the not very useful equality $0=0$.
$$(x_i\,,\,y_i)\;,\,\,i=1,2,3\;,\;\;\text{are collinear}\;\;\iff \frac{y_3-y_2}{x_3-x_2}=\frac{y_3-y_1}{x_3-x_1}=\frac{y_2-y_1}{x_2-x_1}\iff$$
$$\iff (x_2-x_1)(y_3-y_2)=(y_2-y_1)(x_3-x_2)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_1)(y_2-y_1)=(x_2-x_1)(y_3-y_1)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_2)(y_2-y_1)=(x_2-x_1)(y_3-y_2)$$
If one of the denominators vanishes (and thus all of them), then $\,x_1=x_2=x_3\,$ , and we have two columns linearly dependent and the determinant in then zero, otherwise and using the above equality:
$$\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=$$
$$=(x_3-x_2)(y_2-y_1)+x_2\left[(y_3-y_1)-(y_3-y_2)\right]+x_3(y_1-y_2)=$$
$$=0$$
so again zero. Do it slowly...!
Best Answer
Suppose that the equation $$\begin{pmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{pmatrix} \begin{pmatrix}a \\ b \\ c\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$$ has a nontrivial solution $(a,b,c)$. Then all three points $(x_1,y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ lie on the line $ax + by + c =0$.