Where am I going wrong in calculating the projection of a vector onto a subspace?

I am currently working my way through Poole's Linear Algebra, 4th Edition, and I am hitting a bit of a wall in regards to a particular example in the chapter on least squares solutions. The line $$y=a+bx$$ that "best fits" the data points $$(1,2)$$, $$(2,2)$$, and $$(3,4)$$ can be related to the (inconsistent) system of linear equations $$a+b=2$$ $$a+2b=2$$ $$a+3b=4$$ with matrix representation $$A\mathbf{x}=\begin{bmatrix}1&1\\1&2\\1&3\\\end{bmatrix}\begin{bmatrix}a\\b\\\end{bmatrix}=\begin{bmatrix}2\\2\\4\\\end{bmatrix}=\mathbf{b}$$ Using the least squares theorem, Poole shows that the least squares solution of the system is $$\overline{\mathbf{x}}=\left(A^T A \right)^{-1} A^T \mathbf{b}=\left(\begin{bmatrix}3&6\\6&14\\\end{bmatrix}\right)^{-1}\begin{bmatrix}8\\18\\\end{bmatrix}=\begin{bmatrix}\frac{7}{3}&-1\\-1&\frac{1}{2}\\\end{bmatrix}\begin{bmatrix}8\\18\\\end{bmatrix}=\begin{bmatrix} \frac{2}{3}\\1\\\end{bmatrix}$$ so that the desired line has the equation $$y=a+bx=\frac{2}{3} +x$$. The components of the vector $$\overline{\mathbf{x}}$$ can also be interpreted as the coefficients of the columns of $$A$$ in the linear combination of the columns of $$A$$ that produces the projection of $$\mathbf{b}$$ onto the column space of $$A$$ [which the Best Approximation Theorem identifies as the best approximation to $$\mathbf{b}$$ in the subspace $$\mathrm{col}(A)$$]. In other words, the projection of $$\mathbf{b}$$ onto $$\mathrm{col}(A)$$ can be found from the coefficients of $$\overline{\mathbf{x}}$$ by $$\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\frac{2}{3}\begin{bmatrix}1\\1\\1\\\end{bmatrix}+1\begin{bmatrix}1\\2\\3\\\end{bmatrix}=\begin{bmatrix}\frac{5}{3}\\\frac{8}{3}\\\frac{11}{3}\\\end{bmatrix}$$ But when I try to calculate $$\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})$$ directly [taking $$\mathbf{a}_{1}$$ and $$\mathbf{a}_{2}$$ to be the first and second columns of $$A$$, respectively], I get $$\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\left(\frac{\mathbf{a}_{1}\cdot\mathbf{b}}{\mathbf{a}_{1}\cdot\mathbf{a}_{1}}\right)\mathbf{a}_{1}+\left(\frac{\mathbf{a}_{2}\cdot\mathbf{b}}{\mathbf{a}_{2}\cdot\mathbf{a}_{2}}\right)\mathbf{a}_{2}=\left(\frac{\begin{bmatrix}1\\1\\1\\\end{bmatrix}\cdot\begin{bmatrix}2\\2\\4\\\end{bmatrix}}{\begin{bmatrix}1\\1\\1\\\end{bmatrix}\cdot\begin{bmatrix}1\\1\\1\\\end{bmatrix}}\right)\begin{bmatrix}1\\1\\1\\\end{bmatrix}+\left(\frac{\begin{bmatrix}1\\2\\3\\\end{bmatrix}\cdot\begin{bmatrix}2\\2\\4\\\end{bmatrix}}{\begin{bmatrix}1\\2\\3\\\end{bmatrix}\cdot\begin{bmatrix}1\\2\\3\\\end{bmatrix}}\right)\begin{bmatrix}1\\2\\3\\\end{bmatrix}$$ $$=\frac{8}{3}\begin{bmatrix}1\\1\\1\\\end{bmatrix}+\frac{18}{14}\begin{bmatrix}1\\2\\3\\\end{bmatrix}=\begin{bmatrix}\frac{8}{3}\\\frac{8}{3}\\\frac{8}{3}\\\end{bmatrix}+\begin{bmatrix}\frac{9}{7}\\\frac{18}{7}\\\frac{27}{7}\\\end{bmatrix}=\begin{bmatrix}\frac{83}{21}\\\frac{110}{21}\\\frac{137}{21}\\\end{bmatrix}$$ I am quite confident that my calculation is incorrect, for a number of reasons. For example, when I take the component of $$\mathbf{b}$$ orthogonal to $$\mathrm{col}(A)$$ $$\mathrm{perp}_{\mathrm{col}(A)}(\mathbf{b})=\mathbf{b}-\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\begin{bmatrix}2\\2\\4\\\end{bmatrix}-\begin{bmatrix}\frac{83}{21}\\\frac{110}{21}\\\frac{137}{21}\\\end{bmatrix}=\begin{bmatrix}-\frac{41}{21}\\-\frac{68}{21}\\-\frac{53}{21}\\\end{bmatrix}$$ I get a vector that is not perpendicular to either $$\mathbf{a}_{1}$$ or $$\mathbf{a}_{2}$$, indicating that this vector is not in the orthogonal complement of $$\mathrm{col}(A)$$. Can somebody help me identify where I'm going wrong in my attempt to calculate the projection of $$\mathbf{b}$$ onto $$\mathrm{col}(A)$$?

The column space of $$A$$, namely $$U$$, is the span of the vectors $$\mathbf{a_1}:=(1,1,1)$$ and $$\mathbf{a_2}:=(1,2,3)$$ in $$\Bbb R ^3$$, and for $$\mathbf{b}:=(2,2,4)$$ you want to calculate the orthogonal projection of $$\mathbf{b}$$ in $$U$$; this is done by $$\operatorname{proj}_U \mathbf{b}=\langle \mathbf{b},\mathbf{e_1} \rangle \mathbf{e_1}+\langle \mathbf{b},\mathbf{e_2} \rangle \mathbf{e_2}\tag1$$ where $$\mathbf{e_1}$$ and $$\mathbf{e_2}$$ is some orthonormal basis of $$U$$ and $$\langle \mathbf{v},\mathbf{w} \rangle:=v_1w_1+v_2w_2+v_3 w_3$$ is the Euclidean dot product in $$\Bbb R ^3$$, for $$\mathbf{v}:=(v_1,v_2,v_3)$$ and $$\mathbf{w}:=(w_1,w_2,w_3)$$ any vectors in $$\Bbb R ^3$$.
Then you only need to find an orthonormal basis of $$U$$; you can create one from $$\mathbf{a_1}$$ and $$\mathbf{a_2}$$ using the Gram-Schmidt procedure, that is $$\mathbf{e_1}:=\frac{\mathbf{a_1}}{\|\mathbf{a_1}\|}\quad \text{ and }\quad \mathbf{e_2}:=\frac{\mathbf{a_2}-\langle \mathbf{a_2},\mathbf{e_1} \rangle \mathbf{e_1}}{\|\mathbf{a_2}-\langle \mathbf{a_2},\mathbf{e_1} \rangle \mathbf{e_1}\|}\tag2$$ where $$\|{\cdot}\|$$ is the Euclidean norm in $$\Bbb R ^3$$, defined by $$\|\mathbf{v}\|:=\sqrt{\langle \mathbf{v},\mathbf{v} \rangle}=\sqrt{v_1^2+v_2^2+v_3^2}$$.
Your mistake is that you assumed that $$\operatorname{proj}_U\mathbf{b}=\frac{\langle \mathbf{b},\mathbf{a_1} \rangle}{\|\mathbf{a_1}\|^2}\mathbf{a_1}+ \frac{\langle \mathbf{b},\mathbf{a_2} \rangle}{\|\mathbf{a_2}\|^2}\mathbf{a_2}\tag3$$ however this is not true because $$\mathbf{a_1}$$ and $$\mathbf{a_2}$$ are not orthogonal.
Aaand I wasn’t using an orthogonal basis for the subspace. The columns of $$A$$ are linearly independent, which means a least squares solution exists, but they are not orthogonal, which explains why my calculation of the projection of the vector $$\mathbf{b}$$ onto the column space of $$A$$ yielded an incorrect result. Applying the Gram-Schmidt Method to the columns of $$A$$ produces an orthogonal basis for $$\mathrm{col}(A)$$, which can then be used to calculate the projection.