Conditions so that the determinant is zero We have $3$ lines with equations $a_{i1}x+a_{i2}y+a_{i3}=0$, $i=1,2,3$. I want to show that $\det ((a_{ij}))=0$ iff the lines are pairwise parallel  of they have a common point. 
We have that $\det ((a_{ij}))=0$ iff we have a zero row. That would mean that we have linear dependency of the rows, so linear dependency of the lines. That means that some lines are parallel or not? 
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EDIT: 
Suppose that that the determinant is zero, that means that the rows are linear dependent. (The elements of the rows are the coefficients of a line in Hesse normal form, $a_{i1}x+a_{i2}y+a_{i3}=0$.) 
So, we have that for example $(a_{11},a_{12},a_{13})=\kappa\,(a_{21},a_{22},a_{23})+\lambda\,(a_{31},a_{32},a_{33})$. 
I want to show that if the determinant is zero then the lines are pairwise parallel or have an intersection point.
If one of $\kappa$ or $\lambda$ is equal to zero, for example $\kappa$ then we have that $(a_{11},a_{12},a_{13})=\lambda\,(a_{31},a_{32},a_{33})$. This means that the normal vector of the first line $(a_{11}, a_{12})$ is a multiple of the other one, so they are parallel, and so the lines are also parallel, right? But this holds only for two lines, not for all the three, or not? 
Do we have to take also the case that $\kappa, \lambda \neq 0$ ? 
Or is this approach wrong? 
 A: You can suppose that the affine space $\mathbb{R}^2$ is the subset of $\mathbb{R}^3$ on defined by $\{(x,y,1),x,y\in \mathbb{R}\}$. With this identification, 
$D_i:a_{i1}x+a_{i2}y+a_{i3}=0$ is the intersection of $P_i:a_{i1}x+a_{i2}y+a_{i3}z=0$ with $\{(x,y,1)\}$. $P_i$ is the kernel of $l_i(x,y,z)=a_{i1}x+a_{i2}y+a_{i3}z$.  Remark that $P_1\cap P_2\cap P_3$ is the kernel of the linear map defined by $A=(a_{ij})$. $det(a_{ij})=0$ is equivalent to saying that the kernel of $A$ is not trivial, its dimension is 1, or 2. (I assume that at least one line $D_i$ is defined).
If the the dimension of $ker(A)=2$, it is equivalent that $P_1\cap P_2\cap P_3$ is a plan and $P_1=P_2=P_3$ this implies that $D_1=D_2=D_3$.
If the the dimension of $ker(A)=1$. If the intersection of $ker(A)$ and $\{(x,y,1)\}$ is not empty. It is a point which is an element of $D_1\cap D_2\cap D_3$.
Suppose that $dim(ker(A))=1$ $ker(A)\cap\{(x,y,1)\}$ is empty this implies that $ker(A)$ is generated by an element of the form $(a,b,0)$. Let $u_i,v_i$ two distinct point of $D_i$, $u_i-v_i$ is vector of $ker(A)$ since its third coordinate is zero. This implies that $D_1,D_2,D_3$ are pairwise parallel.
