System of Linear Equations ($3\times 6$ matrix, parametric answer) Solve the system
\begin{array}{r@{}r@{}r@{}r@{}r@{}r@{}r@{}r}
x_1 &  - 2 x_2 &  - 2 x_3 &       & + 5 x_5 &  - 4 x_6 & = & -1 \cr
    &          &          & - x_4 & + 4 x_5 & + 2 x_6 & = & 7 \cr 
x_1 &  - 2 x_2 &          &       & + 9 x_5 &  - 6 x_6 & = & 3   
\end{array}
\begin{bmatrix}
x1 \\ x2 \\ x3 \\ x4 \\ x5 \\ x6
\end{bmatrix} 
= __ + t1 __ + t2 __ + t3 __ + t4 __ + t5 __
where t1, ..., t5 are all free variables.
Apologies, I'm still trying to figure out how to use LaTeX and don't know how to make t1,...,t5 italicized, or have the [x1,...,x5] inline.
Anyways, I've reduced the matrix to
\begin{Bmatrix}1 & -2 & 0 & 0 & 9 & -6 & 3 \\
0 & 0 & 1 & 0 & 2 & -5 & 2 \\
0 & 0 & 0 & 1 & -4 & -2 & -7 \\
\end{Bmatrix}
and I know 3 things that DO belong in the blanks.
$$
[x1,...,x6] = \langle 3,0,2,-7,0,0\rangle + t1\langle ,1,0,0,0,0\rangle + t2\langle -9,0,-2,4,1,0\rangle
$$
I'm pretty sure one of them should also be $\langle 6,0,5,2,0,1\rangle$, and then the rest are $\langle 0,0,0,0,0,0\rangle$, however that's incorrect and I'm stuck as to why.
Any help is appreciated, thank you

It looks like it's simply a bug in the row reduction.  I get a different reduced row echelon form:
$$
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 0 & -1 & 4 & 2 & 7 \\ 1 & -2 & 0 & 0 & 
9 & -6 & 3 \\ \end{bmatrix}
\xrightarrow{R_3 \gets R_3-R_1}
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 0 & -1 & 4 & 2 & 7 \\ 0 & 0 & 2 & 0 & 
4 & -2 & 4 \\ \end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 2 & 0 & 4 & -2 & 4 \\ 0 & 0 & 0 & -1 & 
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_1 \gets R_1+R_2}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 2 & 0 & 4 & -2 & 4 \\ 0 & 0 & 0 & -1 & 
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_2 \gets \tfrac{1}{2}R_2}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 1 & 0 & 2 & -1 & 2 \\ 0 & 0 & 0 & -1 & 
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_3 \gets -R_3}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 1 & 0 & 2 & \color{red}{-1} & 2 \\ 0 & 0 & 0 & 1 & 
-4 & -2 & -7 \\ \end{bmatrix}
$$
We have free variables $x_2$, $x_5$, and $x_6$ (since their columns do not have leading entries).  So let's set $x_2=t$, $x_5=u$ and $x_6=v$.  The variables $x_1$, $x_3$, and $x_4$ can be solved in terms of the free variables via:
\begin{align*}
x_1-2t+9u-6v &= 3 \\
x_3+2u-v &= 2 \\
x_4-4u-2v &= -7.
\end{align*}
So the solutions are $$(x_1,x_2,x_3,x_4,x_5,x_6)=(3+2t-9u+6v,t,2-2u+v,-7+4u+2v,u,v)$$ for any $t,u,v$.  We could alternatively write this as
$$(3,0,2,-7,0,0)+t(2,1,0,0,0,0)+u(-9,0,2,4,1,0)+v(6,0,1,2,0,1).$$
