For the following matrix equation: $$\begin{bmatrix}1&1&2\\2&1&3\\3&1&4\\4&1&5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\\4\end{bmatrix}$$
I'm wondering if the collection of all possible $\begin{bmatrix}x\\y\\z\end{bmatrix}$ that solves the above equation has any geometric meaning $R^3$. Would they form a line? A plane? Or nothing at all?
The solution set to the system is the following $$\left\{\begin{pmatrix} 1 \\ 0\\ 0\end{pmatrix}+t\begin{pmatrix} -1 \\ -1\\ 1\end{pmatrix}:\, t\in \mathbb{R}\right\}$$ Indeed, it is the line that passes through $(1,0,0)$, with the same direction as $(-1,-1,1)$.
$\textbf{Edit:}$ (in response to your request) the system can be written as $A\mathbf{x}=b$, where $$A=\begin{pmatrix} 1&1&2 \\ 2&1&3 \\ 3&1&4 \\ 4&1&5 \end{pmatrix} \qquad \mathbf{x}=\begin{pmatrix} x \\ y\\ z\end{pmatrix} \qquad b=\begin{pmatrix} 1 \\ 2\\ 3 \\ 4\end{pmatrix}$$ The augmented matrix of the system is $$(A|\, b)=\begin{pmatrix} 1&1&2&1 \\ 2&1&3&2 \\ 3&1&4&3 \\ 4&1&5&4 \end{pmatrix}$$ and its reduced row echelon form is $$\begin{pmatrix} 1&0&1&1 \\ 0&1&1&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}$$ That is: $$\left\{\begin{align} x+z=1 \\ y+z=0 \end{align}\right.$$ Making $z=t$, we have $x=1-t$ and $y=-t$. Hence any vector solution is in the form $$\begin{pmatrix} x \\ y\\ z\end{pmatrix}=\begin{pmatrix} 1-t \\ -t\\ t\end{pmatrix}=\begin{pmatrix} 1 \\ 0\\ 0\end{pmatrix}+t\begin{pmatrix} -1 \\ -1\\ 1\end{pmatrix}$$