I can not explain a difference between to calculations regarding linear functionals

Hello,
Problem, let $B=${$a_1,a_2,a_3$} be a basis for $C^3$ defined by $a_1=(1,0,-1)$ $a_2=(1,1,1)$ $a_3=(2,2,0)$
Find the dual basis of $B$.

My Solution. Let $W_1$ be the subspace generated by $a_2=(1,1,1)$ $a_3=(2,2,0)$, lets find  $W*$, where $W*$ is the set of linear anihilator of $W_1$. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild $x_2=-x_1$ and $x_3=0$
so from this, by assinging $x_1=1$ ,we find out that the map $f_1=x_1-x_2$.
Similarly,

Let $W_2$ be the subspace generated by $a_2=(1,0,-1)$ $a_3=(2,2,0)$, lets find  $W*$, where $W*$ is the set of linear anihilator of $W_2$. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 0 & -1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild $x_2=-x_1$ and $x_3=x_1$
so from this, by assinging $x_1=1$ ,we find out that the map $f_2=x_1-x_2+x_3$.

Finally,

Let $W_3$ be the subspace generated by $a_2=(1,1,1)$ $a_3=(1,0,-1)$, lets find  $W*$, where $W*$ is the set of linear anihilator of $W_3$. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & -1 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild $x_2=-2x_1$ and $x_3=x_1$
so from this, by assinging $x_1=1$ ,we find out that the map $f_3=x_1-2x_2+x_3$.
So far so Good. The issue is that the last map $f_3$does not agree with the solution I have, but every other thing is the same.Note that I have the final solution with a diffrenet method of solution than mine. In the solution, I got that $f_3=-1/2x_1+x_2-1/2x_3$, I know that I assigned $x_1=1$ when I was trying to find $f_3$, and I could get the same answer if I put $x_1=-1/2$.
This significantly affect the nature of the map. i.e.
let $(5,4,2)\in C^3$, then
$(5,4,2)=1(1,0,-1)+3(1,1,1)+1/2(2,2,0)$
so we have $c_1=1, c_2=3, c_3=1/2$. while
$f_1(5,4,2)=1$,
$f_2(5,4,2)=3$,
$f_3(5,4,2)=-1$
so note that $c_3 \ \ does \ \ not \ \ equal \ \ f_3(5,4,2)$
While if I used the map $f_3=-1/2x_1+x_2-1/2x_3$ we get that $f_3(5,4,2)=1/2$. The equality should ocuur since $f_3$is a vector in the dual which determine the scalar $c_3$.
So my explanation is that I should consider also another equation to each system saying that $c_1x_1+ c_2x_2 + c_2x_3=1$ for each system depending on our choice of the vectors to determine $c_1,c_2,c_3$.
My question, Is this method always work when we find the Dual basis to a given Basis?.
Because I figured this method by myself using a similar method to find anihilator space along with the fact that $dim \ \ W + dim \ \ of \ \ annihilator \ \ space= dim \ \ V$
, where $W\subset V$