import sys

sys.setrecursionlimit(10 ** 6)
from bisect import *
from collections import *
from heapq import *

int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def MF(): return map(float, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LF(): return list(map(float, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
dij = [(0, 1), (1, 0), (0, -1), (-1, 0)]

import numpy as np

def main():
    # 3点Q1,Q2,Q3からできる三角形を底面とし、Pを頂点とする三角すいを考える
    # 底面積(△Q1Q2Q3)と体積が分かれば高さ(点Pと平面Q1Q2Q3の距離)が分かる
    n = II()
    p = np.array(LF())
    qq = [np.array(LF())-p for _ in range(n)]
    ans = 0
    for k, q3 in enumerate(qq):
        for j, q2 in enumerate(qq[:k]):
            for q1 in qq[:j]:
                # 体積vを求める
                v = abs(np.dot(q1, np.cross(q2, q3)) / 6)
                # 底面積sを求める
                s = np.linalg.norm(np.cross(q2 - q1, q3 - q1)) / 2
                # h=3v/sより高さhを求める
                ans += 3 * v / s
    print(ans)

main()