Condition for concurrency of $3$ lines. The question that I am struggling with is : for what value of $k$ do the following $3$ lines have a common point of intersection. 
$$
\begin{split}
x  + y &= 12\\
kx - y &= 0 \\
y  - x &= 2k
\end{split}
$$
This seems like a pretty easy question but I can't figure out how to start working on it. I guess you could probably just start plugging in points but I think there should be a better way to solve this problem. 
Any help would be great.
 A: As explained here, 3 lines are concurrent (have a common point of intersection), when the determinant of the Matrix consisting of the coefficients of each line is zero. Given three lines in the form of ${a}x + {b}y + {c} = 0$, this means:
$\left|  {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}}  \right| = 0$
which expands to
$\Rightarrow {a_3}\left( {{b_1}{c_2} -  {b_2}{c_1}} \right) - {b_3}\left( {{a_1}{c_2} - {a_2}{c_1}} \right) +  {c_3}\left( {{a_1}{b_2} - {a_2}{b_1}} \right) = 0$
A: The foundations of linear algebra contain some theorems to solve this problem. But I assume that you want a more elementary solution (with no matrices, ranks and so on), given that these are only three lines in the plane.
The intersection of the second and third lines will depend on $k$, of course:
$$y=kx$$
$$kx-x=2k$$
$$(k-1)x=2k$$
$$x=\frac{2k}{k-1}\implies y=\frac{2k^2}{k-1}$$
Now, use the first equation:
$$\frac{2k+2k^2}{k-1}=12$$
and solve for $k$.
$$k^2+k=6k-6$$
$$k^2-5k+6=0$$
So $k$ is $2$ or $3$.
A: One trivial approach from first principles: solve for $y$ from the first equation and plug into the other 2, then manipulate $k$ so $x$ comes out the same from both cases.
More advanced approach is with matrices. You are solving a system which results in the following matrix representation:
$$
\begin{pmatrix}
1 &  1 & 12 \\
k & -1 &  0 \\
-1 & 1 & 2k
\end{pmatrix}
\to
\begin{pmatrix}
1 &  1 & 12 \\
k+1 & 0 & 12 \\
0 &  2 & 12+2k
\end{pmatrix}
\to
\begin{pmatrix}
1 & 1 & 12 \\
1 & 0 & \frac{12}{k+1} \\
0 & 1 & 6+k
\end{pmatrix}
\to
\begin{pmatrix}
0 & 0 & 12 - \frac{12}{k+1} - (6+k) \\
1 & 0 & \frac{12}{k+1} \\
0 & 1 & 6+k
\end{pmatrix}
,
$$
assuming $k\ne -1$. If $k=-1$, the second equation has no solution...
Can you finish this?
