Exercise 4.4.15

(a) Gram-Schmidt chooses $q_1=a∕\parallel a\parallel =\frac{1}{3}(1,2,-2)$ and $q_2=\frac{1}{3}(2,1,2)$. Then $q_3=\frac{1}{3}(2,-2,-1)$

(b) The nullspace of $A^T$ contains $q_3$

(c) $\hat{x}={\left(A^{T}A\right)}^{-1}{A}^T(1,2,7)=(1,2)$.