Hint; your final vectors are not correct. The point of GS it to get an orthogonal set of vectors. Are yours orthogonal? You are starting off with two non orthogonal vectors , that is
$v_1=( 1 , 1 , 1)$ and $v_2= ( 1 , 2 ,1)$
The GS algorithm proceeds as follows;
let $w_1=(1,1,1)$
then we define $$w_2= v_2- \frac{\langle v_1 , w_1 \rangle}{\langle w_1 , w_1 \rangle} w_1$$
$$w_2=(1,2,1)-(4/3,4/3,4/3)=(-1/3,2/3,-1/3)$$
and it can be shown now that the set
$$S=\{w_1,w_2\}$$ is orthogonal and also spans the same subspace as the original vectors v.
If we normalize S to say $$S_n=\{(1/3,1/3,1/3),(\frac{-1}{\sqrt6},\sqrt{\frac{2}{3}},\frac{-1}{\sqrt6})\}$$
In general to find the projection matrix P, you first consider the matrix A with your vectors from $S_n$ as columns, that is $$A=\begin{bmatrix} 1/3 & \frac{-1}{\sqrt6} \\ 1/3 & \sqrt{\frac{2}{3}} \\ 1/3 & \frac{-1}{\sqrt6} \\ \end{bmatrix}$$
that is, we will have the orthogonal projection matrix equal to,
$P=A(A^{T}A)^{-1}A^{T}$
Your question is not well clear, but I'm attempting an answer.
First note that the usual definition is that
a linear transformation $P$ is a projector if $P^2=P$
Your additional condition $P^T=P$ caracterize the orthogonal projectors (in a real vector space $V$), where:
$P$ is orthogonal if the kernel of $P$ is orthogonal to its range.
Now note that for any vector $y \in V$ we have that $(y-Py)$ is an element of $\ker(P)$ because $P(y-Py)=Py -P^2y=0$.
So, $P$ is an orthogonal projector if $\forall x,y \in V$ we have :
$$
\langle Px,(y-Py) \rangle=0
$$
But this is done because:
$$
\langle Px,(y-Py) \rangle=\langle P^2x,(y-Py) \rangle=\langle Px,P^T(y-Py) \rangle=\langle Px,(Py-P^2y) \rangle=\langle Px,0 \rangle=0
$$
Best Answer
If $A$ is a $m \times n$ matrix, then $A^T$ is $n \times m$ matrix, and therefore the image of $A^T$ is a subspace of ${\bf R}^n$. It follows that $R(A^T)\oplus R(A^T)^{\perp} = {\bf R}^n$.
Let $x$ be an element of ${\bf R}^n$. Then $x$ can be written uniquely as a sum of an element $y$ of $R(A^T)$ and an element $z$ of $R(A^T)^\perp$.
$(I - Q) x = (I - Q) (y + z) = z \in R(A^T)^\perp$.
Since $x$ was arbitrary, $I - Q$ must be a projection into $R(A^T)^\perp$.
But $R(A^T)^\perp = N(A)$, since if $\langle A^T x, b \rangle = 0$ for all x iff $\langle x, A b \rangle = 0$ for all $x$ iff $A b = 0$.