= Solution
The <linear least-squares problem> is to choose $x\in\mathbb R^n$ minimizing the <Euclidean norm> of the residual:
$$
\boxed{\min_x\|Ax-b\|_2}.
$$
For a <QR decomposition> $A=QR$, orthogonality of $Q$ gives $\|Ax-b\|_2=\|Rx-Q^Tb\|_2$. If $A$ has full column rank and $R=\binom{R_1}{0}$ is in standard form with invertible upper-triangular $R_1\in\mathbb R^{n\times n}$, the minimizer solves $R_1x=(Q^Tb)_{1:n}$.
Applying the <Gram-Schmidt process> to the columns of the given matrix yields
$$
Q=\frac12\begin{pmatrix}
1&1&1&1\\
1&-1&1&-1\\
1&1&-1&-1\\
1&-1&-1&1
\end{pmatrix},
\qquad
R=\begin{pmatrix}
2&1&1\\
0&1&0\\
0&0&1\\
0&0&0
\end{pmatrix}.
$$
Indeed, $Q$ is an <orthogonal matrix> and $A=QR$. Moreover,
$$
Q^Tb=(6,-2,-3,1)^T.
$$
Back substitution in the leading triangular system gives
$$
x_2=-2,
\qquad x_3=-3,
\qquad 2x_1+x_2+x_3=6,
$$
so the unique least-squares solution is
$$
\boxed{x=(11/2,-2,-3)^T}.
$$
The unused final transformed residual component is $1$, so the minimum residual norm is $1$.
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