我们需要找到三个点所在的具有最少顶点数的正多边形。正多边形的顶点均匀分布在圆周上，因此三个点之间的夹角必须是360度的整数分之一。具体步骤如下：

1、计算三个点之间的边长：计算三个点两两之间的距离。 2、计算外接圆半径：使用三角形外接圆公式计算半径。 3、计算中心角：通过余弦定理计算三个点之间的中心角。 4、验证多边形顶点数：检查中心角是否为360度的整数分之一，并找到最小的满足条件的n。

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cstdio>
using namespace std;
 
const double EPS = 1e-6;
const double PI = acos(-1.0);
 
struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
};
 
double dist(Point a, Point b) {
    return hypot(a.x - b.x, a.y - b.y);
}
 
double cross(Point o, Point a, Point b) {
    return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
 
Point circumcenter(Point a, Point b, Point c) {
    double D = 2 * ( (a.x - c.x) * (b.y - c.y) - (a.y - c.y) * (b.x - c.x) );
    double Ux = ( ( (a.x - c.x) * (a.x + c.x) + (a.y - c.y) * (a.y + c.y) ) * (b.y - c.y)
                - ( (b.x - c.x) * (b.x + c.x) + (b.y - c.y) * (b.y + c.y) ) * (a.y - c.y) ) / D;
    double Uy = ( ( (b.x - c.x) * (b.x + c.x) + (b.y - c.y) * (b.y + c.y) ) * (a.x - c.x)
                - ( (a.x - c.x) * (a.x + c.x) + (a.y - c.y) * (a.y + c.y) ) * (b.x - c.x) ) / D;
    return Point(Ux, Uy);
}
 
double angle(Point o, Point a, Point b) {
    double oa = dist(o, a), ob = dist(o, b), ab = dist(a, b);
    double cosTheta = (oa * oa + ob * ob - ab * ab) / (2 * oa * ob);
    return acos(cosTheta);
}
 
int solve(Point a, Point b, Point c) {
    Point o = circumcenter(a, b, c);
    double r = dist(o, a);
    double angle1 = angle(o, a, b);
    double angle2 = angle(o, b, c);
    double angle3 = angle(o, a, c);
 
    vector<double> angles;
    angles.push_back(angle1); 
    angles.push_back(angle2); 
    angles.push_back(angle3); 
 
    for (int n = 3; n <= 200; ++n) {
        bool ok = true;
        for (int i = 0; i < 3; ++i) {
            double theta = angles[i] * n / (2 * PI);
            if (fabs(theta - round(theta)) > EPS) {
                ok = false;
                break;
            }
        }
        if (ok) return n;
    }
    return -1;
}
 
int main() {
    int n;
    cin >> n;
    while (n--) {
        Point a, b, c;
        cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y;
        cout << solve(a, b, c) << endl;
    }
    return 0;
}