#include <iostream>
#include<cmath>
#include<vector>
#include<iomanip>
using namespace std;
const int N = 53;
const double INF = 1e18;
const double eps = 1e-8;
#define zero(x) (((x) > 0 ? (x) : -(x)) < eps)
struct point3
{
    double x, y, z;
    point3 operator+(const point3 &o) const
    {
        return {x + o.x, y + o.y, z + o.z};
    }
    point3 operator-(const point3 &o) const
    {
        return {x - o.x, y - o.y, z - o.z};
    }
    point3 operator*(const double &o) const
    {
        return {x * o , y * o, z * o};
    }
    point3 operator/(const double &o) const
    {
        return {x / o , y / o, z / o};
    }
    bool operator<(const point3 &o) const
    {
        if (!zero(x - o.x))
            return x < o.x;
        if (!zero(y - o.y))
            return y < o.y;
        return z < o.z;
    }
    bool operator!=(const point3 &o) const
    {
        return (!zero(x - o.x) || !zero(y - o.y) || !zero(z - o.z));
    }
} p[N];
vector<point3> line;
double vlen(point3 p)
{
    return sqrt(p.x * p.x + p.y * p.y + p.z * p.z);
}
point3 xmult(point3 u, point3 v)
{
    point3 ret;
    ret.x = u.y * v.z - v.y * u.z;
    ret.y = u.z * v.x - u.x * v.z;
    ret.z = u.x * v.y - u.y * v.x;
    return ret;
}
double dmult(point3 u, point3 v)
{
    return u.x * v.x + u.y * v.y + u.z * v.z;
}
point3 projection(point3 p, point3 u)
{
    return u * dmult(p, u);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i)
        cin >> p[i].x >> p[i].y >> p[i].z;
    if (n <= 3) {
        cout << 0 << '\n';
        return 0;
    }
    double ans = 1e18;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = j + 1; k < n; k++) {
                point3 v1 = xmult(p[j] - p[i], p[k] - p[i]);
                double maxn = -1e18;
                double minn = 1e18;
                for (int l = 0; l < n; l++) {
                    double k = dmult(v1, p[l]) / vlen(v1);
                    maxn = max(maxn, k);
                    minn = min(minn, k);
                }
                ans = min(ans, maxn - minn);
            }
        }
    }
    printf("%.10f\n", ans);
}