$$ v_1 = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} $$ $$ v_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $$ $$ x = \begin{bmatrix} 9 \\ 2 \\ 7 \end{bmatrix} $$
$B = \{v_1, v_2\}$ is a basis for $H = \mathrm{Span}\{v_1, v:2\}$.
Determine if $x$ is in $H$, and if it is, find the coordinate vector $[x]_B$ of $x$ relative to $B$.
So what I started to do was make the Matrix $[v_1 v_2 x]$: $$ [v_1 v_2 x]= \begin{bmatrix} 3&1&9\\ 1&0&2\\ 2&1&7 \end{bmatrix} $$
and then computed the RREF $$ \begin{bmatrix} 1&0&2\\ 0&1&3\\ 0&0&0 \end{bmatrix} $$ This is where I get stuck, because there is zero row on the bottom so there is a free variable. Does this mean that $x$ is NOT in $H$?
Your result means that $$x= 2v_1+3v_2$$
I think you've lost track of what your computations mean. When you set up the matrix $$\begin{bmatrix} 3&1&9\\ 1&0&2\\ 2&1&7 \end{bmatrix}$$
that is shorthand for the system of equations $$ \begin{align} 3a+1b&=9\\ 1a+0b&=2\\ 2a+1b&=7 \end{align}$$ that is to say $$av_1+bv_2 = x$$
When you reduced the matrix to RREF you were effectively solving for $a$ and $b$.
EDIT As for the last couple of sentences in your question, the row of zeros just means that $$0a+0b=0,$$ which is true, though not very interesting. If you had gotten a row like $$0\ \ 0\ \ 1$$ at the bottom, that would mean $$0a+0b=1$$ and since that's nonsense, it would mean that $x$ is not in the span of $v_1$ and $v_2.$