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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 $A$.

Gaussian elimination without pivoting givesThus $\det A=3(\lambda -20)$ and the solution is unique exactly when $\lambda \ne 20$.

For $\lambda =20$, forward substitution in a triangular system in $Ly=b$ gives $y=(1,1,3,\mu -10{)}^T$. Consistency of $Ux=y$ requires $\boxed{\mu =10}$. backward substitution in a triangular system then gives