 Solution

The orthogonal matrix $Q$ preserves the Euclidean norm, soThe first three entries of $Rx-Q^{T}b$ can be set to zero by backward substitution in a triangular system, while its fourth entry is always $3$. Solving the triangular system gives $x_3=16/5$, $x_2=(35-2x_3)/5=143/25$, and $x_1=(98/25-x_2-x_3)/3=-5/3$. Thus the unique least-squares solution and residual norm areUniqueness follows from the nonzero diagonal entries of the top triangular block, equivalently the full column rank of $A$.