What does this condition of concurrency of three lines mean? Let there be three lines given by the equations:
$$a_1x+b_1y+c_1=0\\
  a_2x+b_2y+c_2=0\\
  a_3x+b_3y+c_3=0$$'
Now, a more direct way to prove concurrency of these lines is just proving 
$$\begin{vmatrix}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2\\
a_3 & b_3 & c_3
\end{vmatrix}=0$$ 
But, I came across another condition for proving concurrency which I am not able to understand. It goes as follows:  

Given three lines $$L_1\equiv a_1x+b_1y+c_1=0\\ L_2\equiv
 a_2x+b_2y+c_2=0\\ L_3\equiv a_3x+b_3y+c_3=0$$ are concurrent iff.
  there exist constants $\lambda_1,\lambda_2,\lambda_3$ not all equal to
  zero such that $$\lambda_1L_1+\lambda_2L_2+\lambda_3L_3=0$$ or,
  equivalently:
  $$\lambda_1(a_1x+b_1y+c_1)+\lambda_2(a_2x+b_2y+c_2)+\lambda_3(a_3x+b_3y+c_3)=0$$

What does the last equation mean? What are $x$ and $y$ in it? If $(x,y)$ represent the point of concurrency of the lines, won't every real number $\lambda$ satisfy the above equality? Can someone explain this to me?
 A: For simplicity let's call the new criterion you've found $\mathcal C$.
To say that the three lines are concurrent iff $\mathcal C$ means that starting from the definition of a concurrent system of lines we should be able to derive $\mathcal C$ and starting from $\mathcal C$ we should be able to derive the fact that the lines are concurrent.
Proof that lines are concurrent $\implies \mathcal C$:
Let $$a_1x+b_1y+c_1=0\\
  a_2x+b_2y+c_2=0\\
  a_3x+b_3y+c_3=0$$ such that $\begin{vmatrix}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2\\
a_3 & b_3 & c_3
\end{vmatrix}=0$.  We know that
$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} = 0 \\ \iff \lambda_1(a_1,b_1,c_1)+\lambda_2(a_2, b_2, c_2) + \lambda_3(a_3, b_3, c_3)=0 \text{ has a nontrivial solution}^\dagger$$
That is, iff you can find some numbers $\lambda_1, \lambda_2,\lambda_3$, not all zero, such that each of the following are simultaneously true:
$$\lambda_1a_1+\lambda_2a_2 + \lambda_3a_3 = 0 \\ \lambda_1b_1+\lambda_2b_2 + \lambda_3b_3 = 0 \\ \lambda_1c_1+\lambda_2c_2 + \lambda_3c_3 = 0$$
Now multiply both sides of that first equation by $x$, the second by $y$, and then add them all together to get:
$$(\lambda_1a_1+\lambda_2a_2 + \lambda_3a_3)x + (\lambda_1b_1+\lambda_2b_2 + \lambda_3b_3)y + \lambda_1c_1+\lambda_2c_2 + \lambda_3c_3 = 0$$
Rearranging and factoring out the $\lambda$'s, we get
$$\lambda_1(a_1x+b_1y+c_1) + \lambda_2(a_2x+b_2y+c_2) + \lambda_3(a_3x+b_3y+c_3) = 0 \\ \lambda_1L_1 + \lambda_2L_2 + \lambda_3L_3 = 0$$
Hence the lines are concurrent $\implies \mathcal C$.
I'm not going to do the other implication for you.  See if you can do it yourself.

$\dagger$: Hopefully you've seen this before, but if not you're going to have to learn it.  If you're at a university, see if you can download chapter 2 of this book for more details (specifically look at theorem 2.4).  If you don't have access, see if you can glean out the important facts from the matrix invertibility theorem.
