It is given that,
$\left[\begin{array}{ccc}2 & 1 & 1 \\0 & 3 & -1 \\1 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}1 \\1 \\0\end{array}\right]$
$\Rightarrow \quad 2 x+y+z=1$ $\ldots(\mathrm{i})$
$3 y-z=1$ $\ldots(\mathrm{ii})$
and $\quad x-y+z=0$ $\ldots(\mathrm{iii})$
From Eqs. (i) and (ii), we get $2 x+y+3 y-1=1 \Rightarrow 2 x+4 y=2 \Rightarrow x+2 y=1 \ldots(\mathrm{iv})$
Now, from Eqs. (ii) and (iv), we have
$\frac{x-1}{-2}=y=\frac{z+1}{3}=k$ (let)
$\therefore \quad\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]+k\left[\begin{array}{c}-2 \\ 1 \\ 3\end{array}\right], k \in \mathbf{R}$