Three lines are concurrent (or parallel) $\iff$ the determinant of its coordinates vanishes. I'm trying to prove the concurrency condition for three lines lying on a plane. This condition says that:

Let 
  \begin{cases} ax + by + cz=0 \\ a'x – b'y + c'z=0 \\ a''x + b''y + c''z=0 \end{cases}
  be three lines (barycentric coordinates), with the cordinates of the lines not all equal. Prove that they are concurrent or parallel iff $$ \left| \begin{array}{ccc}
a & b & c \\
a' & b' & c' \\
a'' & b'' & c'' \end{array} \right| =0.
$$

My try:

$$\left| \begin{array}{ccc}
a & b & c \\
a' & b' & c' \\
a'' & b'' & c'' \end{array} \right| =0 \iff \exists P=(x,y,z)\neq(0,0,0),$$
and we've found a common $P$ in those three lines, different of $(0,0,0)$ and with $(x,y)\neq (0,0)$, because it would follow that $P=(0,0,0)$. So the lines are concurrent. 

Is it correct? How I see that they're parallel?
 A: I have also tackled this problem recently. It looks very easy and a basic problem, and yet I haven't been able to convince myself of this proposition.
What I did: 
If the determinant |A| is 0, then the rank of the coefficient matrix
is $\leq 2$.
Thus, the zero space of the given system of equations A x = 0 has dimension at least $\geq 1$.
So, any solution to the system looks like
\[
x'(t) = (tA, B, -1)^{transpose} \quad t \in \mathbb{R}
\]
for some fixed $A, B \in \mathbb{R}$,
if the dimension of the zero space is $1$.
Inserting x' into Ax = 0, we have 
\[
\begin{split}
taA + Bb - c & = 0, \\
ta'A + Bb' - c' & = 0, \\
ta''A + Bb'' - c'' & = 0 .
\end{split}
\]
Solving for B in each relation, we have
 \[
\begin{split}
B & = - taA/b + c/b, \\
B & = - ta'A/b' + c'/b', \\
B & = - ta''A/b'' + c''/b''
\end{split}
\]
for all real $t$, and this is possible if coefficients 
are equal to one another; that is, each relation is of the form
\[
y = a x + c_{i}, \quad i = 1, 2, 3.
\]
If solutions to Ax = 0 look like 
\[
x'(t) = (tA, tB, -1)^{transpose} \quad t \in \mathbb{R},
\]
we can still equate
\[
- taA/b + c/b = - ta'A/b' + c'/b'
\]
and use the same argument concerning coefficients of linear equations.
But I do not think this is correct. Just a suggestion so that anyone
else might correct me.
