= Solution
Let $P=uu^T/\|u\|^2$. This is an <orthogonal projection matrix>, so $P^T=P$ and $P^2=P$. Therefore the <Householder transformation>
$$
H_u=I-2P
$$
satisfies $H_u^TH_u=(I-2P)^2=I$. If $\|a\|=\|b\|$ and $a\ne b$, then
$$
\|a-b\|^2=2(a-b)^Ta,
$$
and hence $H_{a-b}a=a-(a-b)=b$.
Apply a Householder transformation to the first column of $A$ to map it to a multiple of $e_1$, then apply transformations supported on the trailing coordinates to clear each later column below its diagonal. Their product is orthogonal and produces $R$; reversing the product gives $A=QR$.
For the displayed matrix, one resulting factorization is
$$
\boxed{
Q=\frac12\begin{pmatrix}
1&-1&1&1\\
1&1&-1&1\\
1&1&1&-1\\
1&-1&-1&-1
\end{pmatrix},
\qquad
R=\begin{pmatrix}
2&3&2\\
0&5&-2\\
0&0&4\\
0&0&0
\end{pmatrix}.}
$$
Direct multiplication gives the stated $A$, and $Q^TQ=I$.
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