This is to verify if there's an issue with my understanding or if there's issue with the textbook. There seem to be also a previous question here on exactly the same, hoping to help myself and future students.
I did the following and got a different answer to the question, I also verified my answer by testing it for orthogonality against each of it's basis ($u_1 , u_2$ are orthogonal):
Textbook answer:
My working:
$\hat{\vec{y}}$ = projection of $\vec{y}$
$\hat{\vec{y}} = \frac{\vec{y}\cdot \vec{u_1}}{(\vec{u_1})^2} + \frac{\vec{y}\cdot \vec{u_2}}{(\vec{u_2})^2}$
$\hat{\vec{y}} = \frac{\left(\begin{pmatrix}-1\\ \:\:\:\:3\\ \:\:\:\:6\end{pmatrix}\cdot \begin{pmatrix}-5\\ \:\:\:\:-1\\ \:\:\:\:2\end{pmatrix}\right)}{\left(\begin{pmatrix}-5\\ \:\:\:\:\:-1\\ \:\:\:\:\:2\end{pmatrix}\cdot \begin{pmatrix}-5\\ \:\:\:\:\:-1\\ \:\:\:\:\:2\end{pmatrix}\right)}\cdot \begin{pmatrix}-5\\ \:\:\:\:\:-1\\ \:\:\:\:\:2\end{pmatrix}+\frac{\left(\begin{pmatrix}-1\\ \:\:\:3\\ \:\:\:6\end{pmatrix}\cdot \begin{pmatrix}1\\ \:\:\:-1\\ \:\:\:2\end{pmatrix}\right)}{\left(\begin{pmatrix}1\\ \:\:\:\:-1\\ \:\:\:\:2\end{pmatrix}\cdot \begin{pmatrix}1\\ \:\:\:\:-1\\ \:\:\:\:2\end{pmatrix}\right)}\cdot \begin{pmatrix}1\\ \:\:\:-1\\ \:\:\:2\end{pmatrix} = \begin{pmatrix}-1\\ -\frac{9}{5}\\ \frac{18}{5}\end{pmatrix}$
Verify orthogonality:
$\vec{z}=$ vector height of the right angle triangle who's base is the projection ($\hat{\vec{y}}$)
$\vec{z}= \vec{y} - \hat{\vec{y}}=\begin{pmatrix}-1\\ \:3\\ \:6\end{pmatrix}-\begin{pmatrix}-1\\ \:\:-\frac{9}{5}\\ \:\:\frac{18}{5}\end{pmatrix} = \begin{pmatrix}0\\ \frac{24}{5}\\ \frac{12}{5}\end{pmatrix}$
$\vec{z} \cdot \vec{u_1}=0, \vec{z} \cdot \vec{u_2}=0 $
But the textbook answer as the projection is = the vector ($\vec{y}$) itself which by arithmetic doesn't seem to be orthogonal? $\begin{pmatrix}-1\\ \:\:3\\ \:\:6\end{pmatrix}\cdot \begin{pmatrix}-5\\ \:-1\\ \:2\end{pmatrix} = 0 $
The textbook answer is wrong. The only case in which the orthogonal projection of $\vec y$ is $\vec y$ itself is when $\vec y$ belongs to the space onto which we are projecting. That's not the case here. It is possibly a typo; if the third coordinate of $\vec y$ was $-6$ instead of $6$, then, yes, the projection of $\vec y$ would be $\vec y$, because then $\vec y=-\frac13\vec{u_1}-\frac83\vec{u_2}$.