Jtdylktuy

$$
v_1 = begin{bmatrix}
3 \
1 \
2
end{bmatrix}
$$
$$
v_2 = begin{bmatrix}
1 \
0 \
1
end{bmatrix}
$$
$$
x = begin{bmatrix}
9 \
2 \
7
end{bmatrix}
$$

$B = {v_1, v_2}$ 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 $H$?

Note that $$2v_1+3v_2 = x $$ therefore the answer is yes.

$x$ is in the span of $v_1$ and $v_2$

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 $v_2.$