Let $L$ be the line in $\Bbb R^3$ containing the point $A (0,1,1)$.
Specifically, let $L = \{t[0,0,1]+[0,1,1]\mid t\in\Bbb R\}$.
Let $U=[1,-1,0]$
The question is to find the point on the line which is closet to the point $U$.
And here is the answer:
Let $P$ be the point we are looking for.
Then $AU =[1,-2,-1]$.
We see that $AP=\operatorname{proj}_n (AU)$, where $n$ is the normal vector of the line which is $[0,0,1]$.
My question is why $\operatorname{proj}_n (AU)$ equals $AP$. I think it should be $UP$.
In the first sentence, there are many lines containing $A$, but that is fixed in the next sentence. In the next to last sentence, there is not normal vector to $L$. There is a direction vector, which is what you quote.
When you project $AU$ onto $[0,0,1]$ it can't be $UP$ as $P$ is on $L$ and $U$ is not. What you are finding is the distance to move from $A$ along the line until the vector $UP$ is perpendicular to the line.