A linear operator $F : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ projects vectors along (1,2,1) on a subspace My teacher uploaded this question:

And uploaded this answer:

And I have spent way too long trying to understand what is going on. It seem like he is pulling numbers from thin air... :/
I think I got this figured out:

*

*He renamed the components of the plane equation from $x_1, x_2, x_3 $ to a,b,c

*The components of an arbitary vector $u$ is now $u=(x_1,x_2,x_3)$
I understand what is going on visually but I am having some trouble understanding the calculations. $F(u)=u+t_u(1,2,1)$ is this some kind of formula? What does $t_u$ mean? In the next step, the components of $F(u)$ are multiplied and added to the scalars of the plane equation. Why does this work? Is it because it is scaling $F(u)$ the same amount as the plane is scaling itself?
I am overall really close to figuring this out but I need some help.
Unfortunately due to covid all the teachers get bombarded with questions so trying to get an answer before the end of this year is not possible, that's why I am asking here instead of him.
 A: 
We have that $F(u) = u + t_u(1,2,1) = \cdots$ is the projection of $u$ along $(1,2,1)$ on $M$, and therefore satisfies...

This is how your teacher makes sense of "the projection along $(1,2,1)$ onto $M$". For an input point $u = (x_1,x_2,x_3)$, the output $F(u)$ is a specific point on the line through $u$ parallel to $(1,2,1)$. In particular, it is the point on the line where the line intersects the plane $M$.
With that in mind, note that the line through $u$ parallel to $(1,2,1)$ can be parameterized with
$$
r(t) = u + t(1,2,1) = (x_1 + t,x_2 + 2t, x_3 + t).
$$
The output $F(u)$ should be equal to $r(t)$ for some value of $t$ that we will call $t_u$. In other words, we have $F(u) = r(t_u)$. What makes $t_u$ special is that $r(t_u)$ is an element of $M$. Since $r(t_u) = (x_1 + t_u,x_2 + 2t_u, x_3 + t_u)$, $r(t_u)$ is an element of $M$ if
$$
(x_1 + t_u) + 2(x_2 + 2t_u) - (x_3 + t_u) = 0.
$$
Solving this equation for $t_u$ gives us
$$
t_u = -\frac 14 (x_1 + 2x_2 - x_3).
$$
From there, I think the rest should make sense.
