In general, any multiple of an eigenvector is also an eigenvector.
Let $A$ be any matrix and $v$ an eigenvector with eigenvalue $\lambda$. Write $w = \mu v$. We show that $w$ is an eigenvector for the eigenvalue $\lambda$ as follows.
$$
Aw = A(\mu v) = \mu Av = \mu \lambda v = \lambda w
$$
In fact, if the multiplicity of the eigenvalue $\lambda$ is $n>0$ then the number of linearly independent eigenvectors can be anything between $1$ and $n$. For example, tlet $I$ be the $(n\times n)$-identity matrix. Of course, every vector is an eigenvector of $I$ and they all correspond to eigenvalue $1$. So if $n \geq 2$ then $I$ will have multiple linearly independent eigenvectors all corresponding to eigenvalue $1$.
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,
$P=A(A^{T}A)^{-1}A^{T}$
Best Answer
The orthogonal projection onto the one-dimensional subspace spanned by $v$ is $\frac{1}{\| v \|^2}v v^{\top}$. The orthogonal projection onto the complement of $\mathbb{R}v$ is $Id-\frac{1}{\| v \|^2}v v^{\top}$.
The eigenvectors of $Id-\frac{1}{\| v \|^2}v v^{\top}$ are: the vectors of $\mathbb{R}v$ with eigenvalue $0$, and the vectors of the complement of $\mathbb{R}v$ with eigenvalue $1$.
The dimension of $\mathbb{R}v$ is $1$, as $v \neq 0$. The dimension of the complement is $n-1$.