= Solution
The first column is $a=(2,1,2)^T$ with length three. Choose $v=a-3e_1=(-1,1,2)^T$, so $v^Tv=6$. Its <Householder reflection> is
$$
H_1=I-\frac13vv^T=
\begin{pmatrix}2/3&1/3&2/3\\1/3&2/3&-2/3\\2/3&-2/3&-1/3\end{pmatrix}.
$$
Direct multiplication gives
$$
H_1A=\begin{pmatrix}3&3\\0&0\\0&3\end{pmatrix},\qquad
H_1b=\begin{pmatrix}3\\3\\-3\end{pmatrix}.
$$
To eliminate the remaining third-row entry in the second column, reflect the last two coordinates using $v_2=(0,1,-1)^T$. Then
$$
H_2=I-v_2v_2^T=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix},
$$
and
$$
H_2H_1A=\begin{pmatrix}3&3\\0&3\\0&0\end{pmatrix},\qquad
H_2H_1b=\begin{pmatrix}3\\-3\\3\end{pmatrix}.
$$
Both steps are <Householder reflections>; the second happens to exchange two coordinates. Solve the leading triangular equations $3x_1+3x_2=3$ and $3x_2=-3$ to obtain
$$
\boxed{x^*=\begin{pmatrix}2\\-1\end{pmatrix},\qquad
\min_x\|Ax-b\|_2=3.}
$$
For an independent check, $Ax^*-b=(-1,-2,2)^T$, whose squared norm is nine and which is orthogonal to both columns of $A$. Thus $A^T(Ax^*-b)=0$, confirming the least-squares minimizer.
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