The column space of $A$ is $\operatorname{span}\left(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}\right)$.
Those two vectors are a basis for $\operatorname{col}(A)$, but they are not normalized.
NOTE: In this case, the columns of $A$ are already orthogonal so you don't need to use the Gram-Schmidt process, but since in general they won't be, I'll just explain it anyway.
To make them orthogonal, we use the Gram-Schmidt process:
$w_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and $w_2 = \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} - \operatorname{proj}_{w_1} \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$, where $\operatorname{proj}_{w_1} \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$ is the orthogonal projection of $\begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$ onto the subspace $\operatorname{span}(w_1)$.
In general, $\operatorname{proj}_vu = \dfrac {u \cdot v}{v\cdot v}v$.
Then to normalize a vector, you divide it by its norm:
$u_1 = \dfrac {w_1}{\|w_1\|}$ and $u_2 = \dfrac{w_2}{\|w_2\|}$.
The norm of a vector $v$, denoted $\|v\|$, is given by $\|v\|=\sqrt{v\cdot v}$.
This is how $u_1$ and $u_2$ were obtained from the columns of $A$.
Then the orthogonal projection of $b$ onto the subspace $\operatorname{col}(A)$ is given by $\operatorname{proj}_{\operatorname{col}(A)}b = \operatorname{proj}_{u_1}b + \operatorname{proj}_{u_2}b$.
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
Note that the inner product in $V=C([-1,1])$ is defined as $$ \langle f,g\rangle=\int_{-1}^1f(x)g(x)\ dx\tag{1} $$ Now you have three vectors in $V$, $$ v_1=\sin(\pi x), v_2=\sin(2\pi x), v_3=\sin(3\pi x). $$ And of course $f(x)=x$ is another vector in $V$. Denote it as $v=f$. Do you feel more familiar with this setting now and know how to go on?
Moreover, you don't need Gram-Schmidt here since $\{v_1,v_2,v_3\}$ is already an orthonormal set.
Exercise:
Calculate $\langle v,v_i\rangle$ for $i=1,2,3$ using (1). Then the projection is given by $$ a_1v_1+a_2v_2+a_3v_3 $$ where $a_i=\langle v,v_i\rangle$.