Orthogonal projection onto a subspace

Orthonormalization. $u_1 = (1,1,0)/\sqrt2$. The projection of $(0,1,1)$ onto $u_1$ is $\langle (0,1,1), u_1 \rangle u_1 = (1/\sqrt2) \cdot (1,1,0)/\sqrt2 = (1/2)(1,1,0)$. The orthogonal part is
$$(0,1,1) - (1/2)(1,1,0) = (-1/2, 1/2, 1).$$
Norm: $\sqrt{1/4+1/4+1} = \sqrt{3/2}$. So $u_2 = \sqrt{2/3}(-1/2,1/2,1) = (-1/\sqrt6, 1/\sqrt6, \sqrt{2/3})$.

Projection matrix. $P = u_1 u_1^T + u_2 u_2^T$:
$$u_1 u_1^T = \frac12\begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad u_2 u_2^T = \frac16\begin{pmatrix} 1 & -1 & -\sqrt{12}/2 \\ -1 & 1 & \sqrt{12}/2 \\ -\sqrt{12}/2 & \sqrt{12}/2 & 4 \end{pmatrix}.$$
After simplifying (and using $\sqrt{12}/2 = \sqrt3$):
$$P = \frac16\begin{pmatrix} 4 & 2 & -3 \\ 2 & 4 & 3 \\ -3 & 3 & 4 \end{pmatrix}.$$

Verification. $P$ is symmetric, $P^2 = P$, and $P$ fixes $V$ (check on the generators).