Computational Geometry - the ball and the intersection of the straight line

If requested, then the intersection of the line and the ball, and then sentenced to about a point on whether the segment can be.

const double eps=1e-8;
int sgn(double x) { return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1); }
struct P{
    double x,y,z;
    P(double x=0,double y=0,double z=0):x(x),y(y),z(z){
    };
};
P operator-(const P &a,const P&b){
    return P(a.x-b.x,a.y-b.y,a.z-b.z);
}
P operator/(const P&a,double k){
    return P(a.x/k,a.y/k,a.z/k);
}
bool operator==(const P&a,const P&b){
    return a.x==b.x&&a.y==b.y&&a.z==b.z;
}
double dot(const P&a,const P&b){
    return a.x*b.x+a.y*b.y+a.z*b.z;
}
double dist(const P&p){
    return sqrt(p.x*p.x+p.y*p.y+p.z*p.z);
}
vector<P> c_l_intersection(const P&o,const P&s,const P&t,double r){
    vector<P>ret;
    if(s==t) return ret;
    P vec=(t-s)/dist(t-s);
    P onew=o-s;
    double dotov=dot(onew,vec);
    double delta=4*(dotov*dotov-dist(onew)*dist(onew)+r*r);
    delta=sqrt(delta);
    if(sgn(delta)<0) return ret;
    double t1=dotov+delta/2;
    ret.push_back (P (sx + t 1 * vec.x, s.y + t1 * vec.y, t1 + S.E.M. * vec.z));
    
    if(sgn(delta)>0){
        double t2=t1-delta;
        ret.push_back (P (sx + t 2 * vec.x, s.y + t2 * vec.y, t2 + S.E.M. * vec.z));
    }
    Return the right;
}