What does this determinant to test concurrency of three lines mean? I've stumbled upon the determinant way of testing line concurrency which says that, if three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ and $a_3x+b_3y+c_3=0$ are concurrent, then $$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$$
I want to understand this using the concept of vectors. So far I can tell that this is referring to a transformation along the line normals and showing that it will have a 1-D span. 
But what does the matrix have to do with $c_1,c_2,c_3$ since the only $a,b$ define the normal of the line?
 A: The three lines are concurrent iff the system of equations is consistent. This will be the case when the augmented matrix $$A = \left[\begin{array}{cc|c}a_1&b_1&-c_1\\a_2&b_2&-c_2\\a_3&b_3&-c_3\end{array}\right]$$ has the same rank as the $3\times2$ coefficient matrix. The latter has rank at most 2, so a necessary condition is that $\det A=0$. Multiplying a column by $-1$ changes the sign of the determinant, but we only care about its vanishing, hence we can also use the matrix in your question.  
In order to understand this in terms of normals, it might be helpful to make the equations homogeneous by moving to $\mathbb R^3$. The equation $a_ix+b_iy+c_iz=0$ describes a plane through the origin with normal $(a_i,b_i,c_i)^T$. For the three planes to intersect somewhere other than the origin, the homogeneous system’s coefficient matrix must be rank-deficient, which in turn is equivalent to its determinant vanishing. We connect these planes back to lines in $\mathbb R^2$ by intersecting them with the plane $z=1$. (This is one of the standard models of the real projective plane.)
