 Solution

The first two columns are already orthogonal and have norms 15 and 5. Their normalized versions are $q_1=(3,4,0,0{)}^T/5$ and $q_2=(0,0,4,-3{)}^T/5$. Their inner products with the third column are 10 and 5. Subtracting these projections leaves $(20,-15,0,0{)}^T$, whose norm is 25. Thus an economy-size QR factorization isDirect multiplication gives $Q^{T}Q=I_3$ and $QR=A$. If a full square $Q$ is required, append $q_4=(0,0,3,4{)}^T/5$ and append a zero fourth row to $R$; the product is unchanged and the completed $Q$ is an orthogonal matrix.