= Solution
The first column already has zeros below the top entry. In the last three rows of the second column, reflect $w=(4,2,4)^T$ to $(-6,0,0)^T$ using $u=w+6e_1=(10,2,4)^T$. Since $u^Tu=120$, the <Householder reflection> is
$$
H=I-\frac{uu^T}{60}
=\begin{pmatrix}-2/3&-1/3&-2/3\\-1/3&14/15&-2/15\\-2/3&-2/15&11/15\end{pmatrix}.
$$
With $Q=\operatorname{diag}(1,H)$, direct multiplication gives
$$
QA=\begin{pmatrix}1&2\\0&-6\\0&0\\0&0\end{pmatrix},\qquad
Qb=\begin{pmatrix}1\\-2/3\\5/3\\-5/3\end{pmatrix}.
$$
The transformed least-squares objective is $(x_1+2x_2-1)^2+(-6x_2+2/3)^2+50/9$. Its first two squares can both vanish, so the unique answer is
$$
\boxed{x_2=\frac19,\qquad x_1=\frac79,\qquad\min\|Ax-b\|^2=\frac{50}{9}.}
$$
Uniqueness follows because the triangular leading block is invertible; this calculation implements <Householder QR decomposition> without squaring the condition number through normal equations.
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