Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
1 & 2 & 4\\
5 & 11 & 21\\
3 & 7 & 13
\end{bmatrix}$
Obtain: $A=\begin{bmatrix}
1 & 2 & 4\\
5 & 11 & 21\\
3 & 7 & 13
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 & 4\\
0 & -1 & -1\\
0 & -1 & -1
\end{bmatrix}\approx\begin{bmatrix}
1 & 2 & 4\\
0 & -1 & -1\\
0 & 0 & 0
\end{bmatrix}\\
\rightarrow Rowspace(A)=(1,2,4),(0,-1,-1)$
Since the subspace has dimension $2$ we have:
$z(1)+2-4=4-4=0\\
z(0)-1+1=-1+1=0$
Hence, rowspace(A) corresponds to the plane with Cartesian equation $2x + y − z = 0$
$A=\begin{bmatrix}
1 & 2 & 4\\
5 & 11 & 21\\
3 & 7 & 13
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0\\
5 & -1 & -1\\
3 & -1 & -1
\end{bmatrix}\\
\rightarrow Colspace(A)=(1,5,3),(0,-1,-1)$
Thus, this forms the base for the subspace $S=(x,y,z):2x-y+z=0$
$S$ corresponds to all points of $R^3$ in plane $2x-y+z=0$
