I have 2 vectors, a and b, from the origin to a plane.
The normal vector to the plane, is n.
(a dot n) and (b dot n) is the same, and I learned it is equal to the projection of a onto n (and b onto n).
So, this projection should be equal to the length from the origin to the point of intersection of the normal line and the plane.
But it is not equal that! Why ? I have 4 instead of 1.206.
See here : https://www.desmos.com/3d/614b66604f
Things are easier to understand when we have a normal vector with length $1$ (which is what you described in the comment). So let $u=n/|n|$. Taking the dot product of a vector $v$ with $u$ is measuring how much of $v$ is in the direction of $u$, while ignoring the part of $v$ that is orthogonal (perpendicular) to $u$.
More precisely, we let $v_\parallel=(v\cdot u)u$ and $v_\perp = v-v_\parallel$. We use the symbols $\parallel$ and $\perp$ to denote parallel and perpendicular to $u$.
Note that $$v_\parallel \cdot u=(v\cdot u)(u\cdot u)=v\cdot u|u|^2=v\cdot u,$$ so $$v_\perp \cdot u=v\cdot u-v_\parallel\cdot u=0,$$ where $v_\parallel$ is in the direction of $u$ and $v_\perp$ is orthogonal. If we wanted to write these with the initial vector $n$, it would just be replacing the $u$ with $n/|n|$.
Planes to which $u$ is a normal vector are now characterized as sets of the form $$\{x:x\cdot u=c\}$$ for a fixed $c$. That's because if $cu$ is on the plane, then the rest of the plane is obtained by adding perpendicular vectors to $cu$. Each time we add a perpendicular vector $y$ to $cu$, we end up with a vector $x$ which has $x_\parallel = cu$ and $x_\perp =y$.
So even though $a$ and $b$ have different lengths, if they're both on the same plane, $a_\parallel=b_\parallel$, and this is determined by $a\cdot u$ and $b\cdot u$. The difference in lengths comes from the perpendicular parts $a_\perp$ and $b_\perp$, which don't affect $a\cdot u$ and $b\cdot u$.