Solving a linear equation system whose matrix becomes singular when reduced I have a linear equation system:
$$\left\{\matrix{2x_1+x_2-x_3=-1\\3x_1-2x_2+x_3=7\\x_1-3x_2+2x_3=8}\right.$$
During the process of solving it with Gauss-Jordan elimination, I get a singular matrix:
$$\left(\matrix{3&-2&1\\0&7/3&-5/3\\0&0&0}\right.\left|\matrix{7\\-17/3\\0}\right)$$
Converting this matrix to a system:
$$\left\{\matrix{3x_1-2x_2+1x_3=7\\0x_1+{7\over3}x_2-{5\over3}x_3=-{17\over3}\\0x_1+0x_2+0x_3=0}\right.$$
the third row is true, thus the system has infinite solutions - the planes are cut in a specific line, rather than a dot.
My linear algebra book does not elaborate how to find that specific line. According to it, the final result should be:
$$\vec x=\left(\matrix{5/7\\-17/7\\0}\right)+t\left(\matrix{1/7\\-5/7\\1}\right), t\in\mathbb R$$
How do I solve this?

EDIT: If I continue the Gauss-Jordan elimination, I will then get this singular matrix:
$$\left(\matrix{1&0&-1/7\\0&1&-5/7\\0&0&0}\right.\left|\matrix{5/7\\-17/7\\0}\right)$$
which paves way to solve $x_1$ and $x_2$ as:
$$\left\{\matrix{x_1={1\over7}x_3+{5\over7}\\x_2={5\over7}x_3-{17\over7}}\right.$$
so that it can be plugged into $\vec x$: $\left(\matrix{{1\over7}x_3+{5\over7}\\{5\over7}x_3-{17\over7}\\x_3}\right)$.
 A: I have got
$$x_1-3x_2+2x_3=8$$
$$7x_2-5x_3=-17$$
The third equation is the same as second!
A: To obtain the parameterisation of the set of solutions, determine the row reduced echelon form of the augmented matrix:
\begin{align}
\left[\begin{array}{rrr|r}
2&1&-1&-1 \\3&-2&1&7 \\1&-3&2&8
\end{array}\right]&\rightsquigarrow
\left[\begin{array}{rrr|r}
1&-3&2&8 \\ 2&1&-1&-1 \\ 3&-2&1&7 
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrr|r}
1&-3&2&8 \\ 0&7&-5&-17 \\ 0&7&-5&-17 
\end{array}\right]\rightsquigarrow\\[1ex]
\rightsquigarrow
\left[\begin{array}{rrr|r}
1&-3&2&8 \\ 0&7&-5&-17 \\ 0&7&-5&-17 
\end{array}\right]&\rightsquigarrow
\left[\begin{array}{rrr|r}
1&-3&2&8 \\ 0&1&-\frac5 7&-\frac{17}7 \\ 0&0&0&0 
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrr|r}
1&0&-\frac17&\frac 57 \\ 0&1&-\frac5 7&-\frac{17}7 \\ 0&0&0&0 
\end{array}\right]
\end{align}
This final form says the solutions are
\begin{cases}
x_1=\frac57+\frac17 x_3, \\ x_2=-\frac{17}7+\frac 57x_3,
\end{cases}
or, setting $x_3=t$, we have, in vector form:
$$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}\frac57\\-\frac{17}7\\0\end{pmatrix}+t\begin{pmatrix}\frac17\\-\frac{5}7\\1\end{pmatrix}.$$
A: Set one Variable as $t\in\mathbb R$, e.g $x_3=t$ and solve the new system.
