# collinear three points in a linear equation $ax+by+z$

So, there is a plane equation $ax+by+z = 0$. Suppose there are three points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$ that is on the given plane.

Then, $\begin{vmatrix} x & y & z & 1\\x_1 & y_1 & z_1 & 1\\x_2 & y_2 & z_2 & 1\\x_3 & y_3 & z_3 & 1 \end{vmatrix}=0$.

What happens if the three points are collinear?

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Then you can remove the top row and last column and still get determinant zero. –  Gerry Myerson Aug 21 '12 at 1:54
In other words, if you try to recover the equation of the plane from this, you will get $0x+0y+0z=0$. –  Rahul Aug 21 '12 at 2:09

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 $0=0$.