Prove a 3x3 system of linear equations with arithmetic progression coefficients has infinitely many solutions How can I prove that a 3x3 system of linear equations of the form:
$\begin{pmatrix}
a&a+b&a+2b\\
c&c+d&c+2d\\
e&e+f&e+2f
\end{pmatrix}
\begin{pmatrix}
x\\ y\\ z
\end{pmatrix}
=\begin{pmatrix}
a+3b\\
c+3d\\
e+3f
\end{pmatrix}$
for $a,b,c,d,e,f \in \mathbb Z$ will always have infinite solutions and will intersect along the line 
$ r=
\begin{pmatrix}
 -2\\3\\0
\end{pmatrix}
+\lambda
\begin{pmatrix}
 1\\-2\\1
\end{pmatrix}$

First, consider the homogeneous system
$$\left(\begin{array}{ccc}
a & a+b & a+2b\\\
c & c+d & c+2d\\\
e & e+f & e+2f
\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right).$$
If $(a,c,e)$ and $(b,d,f)$ are not scalar multiples of each other, then the coefficient matrix has rank $2$, so the solution space has dimension $1$. The vector $(1,-2,1)^T$ is clearly a solution, so the solutions are all multiples of $(1,-2,1)^T$. That is, the solutions to the homogeneous system are $\lambda(1,-2,1)^T$ for arbitrary $\lambda$.
Therefore, the solutions to the inhomogeneous system are all of the form $\mathbf{x}_0 + \lambda(1,-2,1)^T$, where $\mathbf{x}_0$ is a particular solution to this system. Since $(-2,3,0)$ is a particular solution always, then all solutions have the described form.
If one of $(a,c,e)$ and $(b,d,f)$ is a multiple of the other, though, then there are other solutions: the matrix has rank $1$, so the nullspace has dimension $2$. Say $(a,c,e) = k(b,d,f)$ with $k\neq 0$, then there is another solution: $(-1-\frac{1}{k},1,0)$ would also be a solution to the system, so that the solutions to the inhomogeneous system would be of the form
$$r = \left(\begin{array}{r}-2\\3\\0\end{array}\right) + \lambda\left(\begin{array}{r}1\\-2\\1\end{array}\right) + \mu\left(\begin{array}{r}-1-\frac{1}{k}\\1\\0\end{array}\right).$$
This includes the solutions you have above, but also others. (If $k=0$, then you can use $(0,-2,1)$ instead of $(-1-\frac{1}{k},1,0)$) 
If $(b,d,f)=(0,0,0)\neq (a,c,e)$, then $(1,0,-1)$ can be used instead of $(-1-\frac{1}{k},1,0)$ to generate all solutions.
And of course, if $(a,b,c)=(b,d,f)=(0,0,0)$, then every vector is a solution.
In all cases, you have an infinite number of solutions that includes all the solutions you give (but there may be solutions that are not in that line).
