Deriving equation of plane from reduced row echelon form Say I reduced a particular matrix to its reduced row echelon form and have this: 
$$
\begin{bmatrix}
1 & 0 & 5 & 0\\ 
0 & 1 & 3 & 0\\ 
0 & 0 & 0 & -2
\end{bmatrix}
$$
And I let $\vec{b}=
\begin{bmatrix}
x\\ 
y\\ 
z
\end{bmatrix}$ so that $Ak=b$ looks like this:
$$
\begin{bmatrix}
1 & 0 & 5 & 0\\ 
0 & 1 & 3 & 0\\ 
0 & 0 & 0 & -2
\end{bmatrix}
\begin{bmatrix}
k_{1}\\ 
k_{2}\\ 
k_{3}\\ 
k_{4}
\end{bmatrix}=\vec{b}
$$
Now, I can see that the column space of this matrix $A$ is likely to be a plane. But how do I find the equation of the plane, without using cross product?
I could get to something like:
$-2k_{4}=z$
$k_{2}+3k_{3}=y$
$k{1}+5k_{3}=x$
But this is still far from the equation of the plane. I don't intend to use cross product because I am thinking if I could extend this to higher dimensions which cross product could be very cumbersome.
So, how do I derive the equation of the plane of the column space of matrix A from here?
Thanks for any help! 
 A: You seem to be confusing the solution set of $A\mathbf{x}=\mathbf{b}$ with the column space of $A$.
Remember: the system $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $\mathbf{b}$ is in the column space. But the solutions themselves (the $\mathbf{x}$ that satisfy the equation) are not in the column space generally. Here, they don't even "live" in the same space, as the column space of $A$ is a subspace of $\mathbb{R}^3$, but the solutions are vectors in $\mathbb{R}^4$.
Here, the column space of $A$ has dimension $3$: the column space includes the vectors
$$\left(\begin{array}{c}1 \\0\\0\end{array}\right),\quad \left(\begin{array}{c}0\\1\\0\end{array}\right),\quad\text{and}\quad \left(\begin{array}{r}0\\0\\-2\end{array}\right),$$
which clearly span all of $\mathbb{R}^3$. So the column space is not a plane, it is all of $\mathbb{R}^3$.
I think your confusion lies in the following: the number of parameters (free variables/non pivot columns) determines the dimension of the solution space to $A\mathbf{x}=\mathbf{b}$ when it has a solution. And in order to decide whether $A\mathbf{x}=\mathbf{b}$ has a solution, you look at the column space. The system $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $\mathbf{b}$ lies in the columnspace of $A$. 
Because here the column space is all of $\mathbb{R}^3$, the system has solutions for every possible choice of $\mathbf{b}\in\mathbb{R}^3$. 
The solution has dimension $\mathrm{nullity}(A)$, which by the Rank-Nullity Theorem is equal to the number of columns of $A$ minus the rank of $A$, i.e., the number of free variables/number of parameters/number of non pivot columns in $A$. Here, the rank is $3$, so the nullity is $1$. That is: the solution space has dimension $1$, and so will be a line in $\mathbb{R}^4$.
Then, to get the equation of that line, we have
$$\begin{align*}
k_1 +5k_3 &= x\\
k_2 + 3k_3 &= y\\
k_4 &= -\frac{1}{2}z.
\end{align*}$$
Note that $x$, $y$, and $z$ are constants (they are the entries of your $\mathbf{b}$, and the coordinates in $\mathbb{R}^4$ (where the line "lives") are given by components called $k_1$, $k_2$, $k_3$, and $k_4$. 
The system of lines in $\mathbb{R}^4$ that are solutions to the different systems are parametrized by your $x$, $y$, and $z$, but since they can be any point in $\mathbb{R}^3$, there is no relation among them; the collection of all $\mathbf{b}$s for which the system is consistent is all of $\mathbb{R}^3$, not a plane either. 
