QR factorization by Gram-Schmidt

Columns. $a_1 = (1,1,0)$, $a_2 = (1,0,1)$.

First vector. $q_1 = a_1 / \|a_1\| = (1,1,0)/\sqrt2$.

Projection. $\langle a_2, q_1 \rangle = (1\cdot1 + 0\cdot1 + 1\cdot0)/\sqrt2 = 1/\sqrt2$. The orthogonal part is
$$a_2 - \langle a_2, q_1 \rangle q_1 = (1,0,1) - \frac12(1,1,0) = (1/2, -1/2, 1).$$
Normalize: $\|a_2 - \langle a_2, q_1 \rangle q_1\| = \sqrt{1/4+1/4+1} = \sqrt{3/2}$. So
$$q_2 = \sqrt{2/3}\,(1/2, -1/2, 1) = (1/\sqrt6, -1/\sqrt6, \sqrt{2/3}).$$

Matrices $Q$ and $R$.
$$Q = \begin{pmatrix} 1/\sqrt2 & 1/\sqrt6 \\ 1/\sqrt2 & -1/\sqrt6 \\ 0 & \sqrt{2/3} \end{pmatrix}, \qquad R = \begin{pmatrix} \sqrt2 & 1/\sqrt2 \\ 0 & \sqrt{3/2} \end{pmatrix}.$$

Conclusion. $A = QR$ with $Q$ having orthonormal columns and $R$ upper triangular.