#include <bits/stdc++.h>
#ifdef dbg
#define D(...) fprintf(stderr, __VA_ARGS__)
#define DD(...) D(#__VA_ARGS__ " = "), debug_helper::debug(__VA_ARGS__), D("\n")
#include "C:\Users\wsyear\Desktop\OI\templates\debug.hpp"
#else
#define D(...) ((void)0)
#define DD(...) ((void)0)
#endif
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;

using namespace std;

const int N = 55;

struct coor
{
    int x,y,z;//坐标，也可以表示成向量，向量也是一个坐标表示嘛
    coor(){}
    coor(int xx,int yy,int zz):x(xx),y(yy),z(zz){}
    bool operator &(coor a) const //判断两个向量是否共线，共线返回true
    {
        //思路：判断叉积，是否各项系数都是0
        return (y*a.z - z*a.y)==0 && (z*a.x - x*a.z)==0 && (x*a.y - y*a.x)==0;
    }
    coor operator ^(coor a) const //得到两个向量的叉积（就是向量积），返回的是一个向量（坐标）
    {
        return coor(y*a.z - z*a.y,z*a.x - x*a.z,x*a.y - y*a.x);
    }
    coor operator -(coor a) const //如果是c-d的话，得到向量dc，
    {
        return coor(x-a.x,y-a.y,z-a.z);
    }
    int operator *(coor a) const //得到两个向量的 数量积，返回整数即可
    {
        return x*a.x+y*a.y+z*a.z;
    }
}zb[N];

bool all_in_Aline(int n)
{
    //思路，暴力枚举，每三个点，看看是不是所有叉积都是0
    for (int i=1;i<=n;++i) //这个作为起点吧
        for (int j=i+1;j<=n;++j)
            for (int k=j+1;k<=n;++k)
            {
                coor t1 = zb[k]-zb[i];
                coor t2 = zb[j]-zb[i];
                if (t1&t2) continue;
                return false;
            }
    return true;
}

bool checkThree (coor a,coor b,coor c)
{
    return (b-a)&(c-a);
}

int n, a[N], b[N], c[N];
mt19937 rnd(random_device{}());

template<class T> double pw(T x) { return 1. * x * x; }
template<class T> int sign(T x) { return x == 0 ? 3 : (x < 0 ? 1 : 2); }

double calc(int x, int y, int z) {
  if (checkThree(zb[x],zb[y],zb[z])) return 1e18;
  coor p1 = zb[x], p2 = zb[y], p3 = zb[z];
  double A = ( (p2.y-p1.y)*(p3.z-p1.z)-(p2.z-p1.z)*(p3.y-p1.y) );
  double B = ( (p2.z-p1.z)*(p3.x-p1.x)-(p2.x-p1.x)*(p3.z-p1.z) );
  double C = ( (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x) );
  double D = ( 0-(A*p1.x+B*p1.y+C*p1.z) );
  int And = 3;
  double mx = 0, ss = sqrt(pw(A) + pw(B) + pw(C));
  rep (i, 1, n) {
    And &= sign(A * a[i] + B * b[i] + C * c[i] + D);
    mx = max(mx, abs(A * a[i] + B * b[i] + C * c[i] + D) / ss);
  }
  return And ? mx : 1e18;
}

int main() {
  cin.tie(nullptr) -> ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  cin >> n;
  rep (i, 1, n) cin >> a[i] >> b[i] >> c[i];
  if (n == 1) return cout << double(0), 0;
  if (n == 2) return cout << sqrt(pw(a[1] - a[2]) + pw(b[1] - b[2]) + pw(c[1] - c[2])), 0;
  rep (i, 1, n) zb[i] = (coor){a[i], b[i], c[i]};
  if (all_in_Aline(n)) return cout << double(0), 0;
  double ans = 1e18;
  rep (i, 1, n) rep (j, i + 1, n) rep (k, j + 1, n) ans = min(ans, calc(i, j, k));
  cout << (ans > 1e17 ? 0. : ans) << '\n';
}