There is a standard derivation of the equation of a plane, when the following quantities are given
$1.)$ A vector normal to the plane, from the origin $:$ $\vec{ON}$ $=$ $d{ô}$, where $d$ is some real number, and $d≠0$
$ô$ is the unit vector in the direction of $\vec{ON}$
$2.)$ An arbitrary point $P$ on the plane with the position vector $\vec{r}$
We know that $\vec{ON}.\vec{NP}=0$
From this follows: $(\vec{r}-dô).dô=0$
In my textbook, after solving this equation, it is shown that since $(\vec{r}-dô).dô=0$
$(\vec{r}-dô).ô=0$
$\vec{r}ô-d=0$
$\vec{r}ô= d$
I was wondering about any other approach from the equation $(\vec{r}-dô).dô=0$ to get an equation that is equivalent of the equation $\vec{r}.ô=d$
I tried in the following manner
$\vec{r}.dô-dô.dô=0$
$\vec{r}.dô-d^2=0$
$\vec{r}.dô=d^2$
How is this equation related to $\vec{r}ô= d$
We can also state that how to prove that $(\vec{r}.dô)=(\vec{r}.ô)d$
If I understand your notation correctly, $d$ is just some real number. The inner product you are using in order to express orthogonality is what one calls bilinear, i.e. (using your (non-standard) notation) $$(a \vec{v}_1+b \vec{v}_2).\vec{v}_3 = a(\vec{v}_1.\vec{v}_3)+b(\vec{v}_2.\vec{v}_3)$$ for any choice of vectors $\vec{v}_1,\vec{v}_2,\vec{v}_3\in\mathbb{R}^n$ and $a,b\in\mathbb{R}$. So coming back your first question, one has
$$ \vec{r}.dô=d^2 \Leftrightarrow d(\vec{r}.ô) = d^2 \Leftrightarrow \vec{r}.ô = d,$$ which is your original equation. Using above notion of bilinearity, I think you can figure out the answer to your second question by yourself.