How can an almost arbitrary plane in a 3D dataset be plotted by matplotlib?

Question

There is an array containing 3D data of shape e.g. (64,64,64), how do you plot a plane given by a point and a normal (similar to hkl planes in crystallography), through this

Answer 1

I have the penultimate solution for this problem. Partially solved by using the second answer to Plot a plane based on a normal vector and a point in Matlab or matplotlib :

# coding: utf-8
import numpy as np
from matplotlib.pyplot import imshow,show

A=np.empty((64,64,64)) #This is the data array
def f(x,y):
    return np.sin(x/(2*np.pi))+np.cos(y/(2*np.pi))
xx,yy= np.meshgrid(range(64), range(64))
for x in range(64):
    A[:,:,x]=f(xx,yy)*np.cos(x/np.pi)

N=np.zeros((64,64)) 
"""This is the plane we cut from A. 
It should be larger than 64, due to diagonal planes being larger. 
Will be fixed."""

normal=np.array([-1,-1,1]) #Define cut plane here. Normal vector components restricted to integers
point=np.array([0,0,0])
d = -np.sum(point*normal)

def plane(x,y): # Get plane's z values
    return (-normal[0]*x-normal[1]*y-d)/normal[2]

def getZZ(x,y): #Get z for all values x,y. If z>64 it's out of range
    for i in x:
        for j in y:
            if plane(i,j)<64:
                N[i,j]=A[i,j,plane(i,j)]

getZZ(range(64),range(64))
imshow(N, interpolation="Nearest")
show()

It's not the ultimate solution since the plot is not restricted to points having a z value, planes larger than 64 * 64 are not accounted for and the planes have to be defined at (0,0,0).