Please see the image with the question and my answer. The thing I'm having an issue with is understanding what this vector description actually is. Is it a 2-dimensional flat in $\mathbb{R}^6$?
In fact, according to the table, we have
$n-2$ Cartesian equations and $2$ free parameters in this example. So since $n−2=4$ then $n=6$? Is this true?
Thank you for taking the time to look at this post. Really appreciate your help.
The table constructs a formula between the number of free parameters $F$, number of equations $E$, and the dimension of the space $n$ as follows
$$n-F=E\tag{1}$$
however, it should be emphasized that $E$ is the number of linearly independent equations and not just the number of equations! Now, let us see how many linearly independent equations we have in your example. For doing so we may perform the usual elementary row operations on the system of equations. This can easily be done with the augmented matrix
\begin{align} \left[ \begin{array}{ccccc|c} 1 & 0 & 3 & 2 & 0 & 8\\ 0 & 0 & 1 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 1 & 2\\ 3 & 0 & 8 & 5 & -3 & 18 \end{array} \right] &\to \left[ \begin{array}{ccccc|c} 1 & 0 & 3 & 2 & 0 & 8\\ 3 & 0 & 8 & 5 & -3 & 18 \\ 0 & 0 & 1 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 1 & 2\\ \end{array} \right] \\ &\to \left[ \begin{array}{ccccc|c} 1 & 0 & 3 & 2 & 0 & 8\\ 0 & 0 & \frac{1}{3} & \frac{1}{3} & 1 & 2 \\ 0 & 0 & 1 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 1 & 2\\ \end{array} \right] \\ &\to \left[ \begin{array}{ccccc|c} 1 & 0 & 3 & 2 & 0 & 8\\ 0 & 0 & \frac{1}{3} & \frac{1}{3} & 1 & 2 \\ 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3}\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{array} \right] \\ \end{align}
So as we can see there are just $3$ linearly independent equations in your example as the last row turned out to be zero. So in your example we will get
\begin{align} F&=2 \\ E&=3 \\ n&=F+E=5 \end{align}
The formula $(1)$ is a simple manifestation of the rank-nullity theorem which you may study in a course dealing with theoretical aspects of linear algebra.