# Number of Planes in 3-D unit distance away from three given non-collinear points are?

Number of Planes in 3-Dimensional geometry unit distance away from points $A(3,5,1)$, $B(-3,-5,1)$, $C(10,-2,5)$ is ?

I have easily counted two such planes $P_1,P_2$ which are parallel to the plane($P_3$)containing the three points lying on either sides of $P_3$ but the answer given is $8$ and I have no idea about the other $6$ planes . Any Help ?

This is the scene:

The three points $A$, $B$, $C$ and the unit spheres around. Then the plane through the three points.

It is clear that the two parallel planes in distance $\pm 1$ to that plane are among the sought ones.

I would agree with David that this is mostly a decision per sphere, if the plane touches above or below (relative to the containing plane). Which seems to lead to eight choices.

We can describe an arbitray plane as $$n \cdot x = d$$ where $n$ is a unit normal vector of the plane and $d$ is the (signed) distance of the plane to the origin.

The spheres are described by $$(x - P)^2 = 1$$ The vectors $x$ within the plane must not enter the spheres: $$(x - P)^2 \ge 1$$ At three points the plane touches the spheres: $$n \cdot x_P = d \quad\quad (x_P - P)^2 = 1 \\$$ So we have the system $$E: n \cdot x = d \\ (x - A)^2 \ge 1 \\ (x - B)^2 \ge 1 \\ (x - C)^2 \ge 1 \\ n \cdot x_A = d \quad\quad (x_A - A)^2 = 1 \\ n \cdot x_B = d \quad\quad (x_B - B)^2 = 1 \\ n \cdot x_C = d \quad\quad (x_C - C)^2 = 1$$ for all $x \in E$ and $13$ real unknowns $n = (n_1, n_2, n_3)^\top$, $d$ and $x_A$, $x_B$, $x_C$.

Create three unit balls centered on the points. Then notice that for the three balls, there is a tangent plane:

• touching the "bottom" of each sphere
• touching the "top" of each sphere
• touching the "top" of sphere $A$ and the "bottom" of $B$ and $C$
• touching the "top" of sphere $B$ and the "bottom" of $A$ and $B$

etc.

$2^3 = 8$.

• A bit difficult to Imagine initially but I think I get what you are talking about. Thanks Dec 19, 2016 at 2:10