If you have $n=72$ points, then you have $n-1=71$ pairs going to any particular point and ${n\choose2}=2485$ total pairs to check. But it's probably easier to just decide which side of the plane each point is on, with an array, or map, $ \eqalign{ \phi &: \{1,~\dots,~n\}\to\{\pm1\} \\\\ \phi &: i \mapsto \phi(i)= \operatorname{sign}(\overrightarrow{PP_i}\cdot\overrightarrow{N}) \\\\ \overrightarrow{PP_i} \cdot \overrightarrow{N} &= a(x_i-x_0)+b(y_i-y_0)+c(z_i-z_0) \\\\ P_i &= (x_i,y_i,z_i)\qquad 0\le i\le n \\\\ \overrightarrow{N} &= (a,b,c) } $ where $P_0$ is the point in the plane, $P_i$ is the $i$th surface/object point, and the normal vector $\overrightarrow{N}$ to the plane defines the direction or side of the plane which $\phi$ maps to $+1$. Then, assuming no points lie in the plane, the line connecting a given pair of points (say $i$ and $j$) crosses the plane iff \phi(P_i)\phi(P_j)<0. This takes $3n$ multiplications and $3n+2n=5n$ additions ($3$ subtractions & $2$ additions for each point) to compute and requires $n$ bits (assuming no points lie directly in the plane) or tertiary ($+1,0$ or $-1$) quantities (if a point can lie in the plane) of storage. For "real world" applications, you might also want to perform $n$ comparisons of the absolute value of the dot products against some $\epsilon$ as a threshold for considering the point to be in the plane rather than naively mixing your machine's representation and rounding errors. Is this a helpful start, too obvious, or irrelevant?