In the real vector space $\mathbb R^3$, equipped with the dot product, I have a vector $V=(a,b,c)$ and a point $P=(x_1, y_1, z_1)$. I need to first find plane that is perpendicular to vector $V$ and passed throught the point $P$. Then on this plane I need to find a point that is at a distance $r$ from the point $P$. I guess, the solution will be a parametric one, since all such equidistant points will form a circle.
Edit: Giving below an illustration. I know $P$, $V$ and $r$ and I need to find any point on the circle with the radius $r$ which lies on the plane that's perpendicular to $V$ and passed includes $P$.
Note that $V\neq0$. There are two cases to consider: either $P\in \mathbb R V$ or $P \notin \mathbb R V$. Otherwise written either V belongs to the vector line (line passing through $0$) directed by V, or V does not belong to the vector line directed by $V$.
(first case) I will take an example: $V=(1,1,1)$ and $P=(2,2,2)$. Let $M=(x,y,z)\in \mathbb R^3$ and $\mathcal{P}$ the plane perpendicular to $V$ and passing through P. $M\in \mathcal{P}\iff \vec{PM}.V=0 \iff (M-P).V=0\iff x-2+y-2+z-2=0 \iff x+y+z=6$ For example, $P'=(5,0,1)\in\mathcal{P}$. Then $A=P+\frac{r}{\|P'-P`\|}(P'-P)$ is a point at the distance $r$ from $P$.
(second case) Let $P'$ be the orthogonal projection of $P$ on $\mathbb RV$ : $P'=\frac{P.V}{\|P\|^2}P$. Then $A=P+\frac{r}{\|P'-P\|}(P'-P)$ is a point a the distance $r$ from $P$.