#include<bits/stdc++.h>
#define SZ(x) ((int)x.size())
#define uni(x) sort(all(x)),x.resize(unique(all(x))-x.begin());
#define GETPOS(c,x) (lower_bound(all(c),x)-c.begin())
#define lown1(x,val) low(in(x),val)-x
#define lowm1(x,val) low(im(x),val)-x
#define low1(x,nums,val) low(x+1,x+nums+1,val)-x
#define mst(x,val) memset((x),val,sizeof((x)))
#define ls rt<<1
#define rs rt<<1|1
#define lson rt<<1,l,M
#define rson rt<<1|1,M+1,r
#define PI acos(-1)
#define MM int M=(l+r)>>1;
#define fu(i,r,t) for(int i=r;i<=t;i++)
#define fd(i,r,t) for(int i=r;i>=t;i--)
#define fh(i,be,e) for(int i=head[be];~i;i=e[i].next)
#define fa(i,V) for(auto i:V)
#define far(i,V) for(auto &i:V)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b))/__gcd(a,b)
#define cp(i,ans) printf("%.if",ans);
#define cpp(i,ans) cout<<setprecision(i)<<fixed<<ans<<endl;
#define ppb pop_back
#define ppf pop_front
#define pb push_back
#define pf push_front
#define pq priority_queue
#define lowbit(x) ((x)&(-x))
#define all(V) V.begin(),V.end()
#define ms multiset
#define mod(x) (((x)<0)?(x)%mo_num+mo_num:(x)%mo_num)
#define vc vector
#define vct vector<int>
#define out(i) cout<<(i)<<endl;
#define fi first
#define se second
#define fun(i) fu(i,1,n)
#define fut(i) fu(i,1,t)
#define fum(i) fu(i,1,m)
#define ld long double
#define umap unordered_map
#define Umap unordered_map<int,int>
#define P pair<int,int>
#define SET set<int>
#define mk make_tuple
#define eps 1e-6
//Remember cancel"#define endl '\n'" in interactive questions or use "<<flush"
#define endl '\n'
#define low lower_bound
#define upp upper_bound
#define yn(key) out(key?"YES":"NO")
//#define yn(key) out(key?"Yes":"No")
#define in(i) i+1,i+1+n
#define im(i) i+1,i+1+m
#define ik(i,k) i+1,i+1+k
#define bffs(i) __builtin_ffs(i)
#define bcount(i) __builtin_popcount(i)
#define bone(i) ((1<<i)-1)
#define db double
#define ll long long
#define got(container,num) get<num-1>(container)
#define int long long
#define print(a,n) fun(i)cout<<a[i]<<(i!=n?' ':endl);
#define outcase(x) cout<<"Case #"<<(++case_of_T)<<": "<<(x)<<endl;
#define ptcase(x) printf("Case #%d: %d\n",++case_of_T,x);
#define plcase(x) printf("Case #%lld: %lld\n",++case_of_T,x);
using namespace std;
//Remember to cancel the line below and declare INT=INT_MAX/2; when you want to change long to int
const int INF=LLONG_MAX/4,SINF=0x3f3f3f3f,Lim=1<<20,MINF=LLONG_MAX;
//const int INF=INT_MAX/4,SINF=0x3f;
// use C:printf("%.16f", x);  ->  printf("%.10f", x); can accelerate the program
const int dx[]={0,0,-1,1},dy[]={-1,1,0,0};//down up left right
const int maxn=1e6+1e5;
int mo_num=1e9+7;
//const int mo_num=998244353;
int n,m,t,a[maxn],b[maxn],ans,case_of_T;

void solve()
{
    int A,B;
    cin>>n>>A>>B;
    if(n<A+B)out(0)
    else{
        m=n+2-A-B;
        int C=mod(m*(m-1)/2);
        ans=mod(4*mod(mod((n-A+1)*(n-B+1))*C));
        ans=mod(ans-mod(4*C*C));
        out(ans)
    }
    return ;
}
main()
{
    IOS

    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

//面倒くさがらずに実験したらすぐ分かった…
//all - NG * NGの形にして、1次元の問題でNGを数える。
//A, Bを置く位置x, y(0≦x, y≦N-1)を考えると
//0 ≦ x ≦ N - A
//0 ≦ y ≦ N - B
//y - A + 1 ≦ x ≦ y + B - 1
//なる(x, y)を数えたくなるので、そういう(x, y)を表示してみる。(実験)
//A + B > Nのときは、(N + 1 - A) * (N + 1 - B)個全てが埋まる.
//A + B <= Nのとき、
//1辺len = N + 1 - A - Bの二等辺三角形2つを引けばよい.
//ただし、二等辺三角形の面積は1 + … + lenなことに注意（離散的なので）
//(N + 1 - A) * (N + 1 - B) - (len + 1) * len個.

#include <iostream>
#define int long long
using namespace std;

int mod = 1000000007;

signed main() {
	int T;
	cin >> T;
	
	while (T--) {
		int n, a, b;
		cin >> n >> a >> b;
		
		int all = (n - a + 1) * (n - a + 1) % mod * (n - b + 1) % mod * (n - b + 1) % mod;
		int ng;
		if (a + b > n) ng = (n + 1 - a) * (n + 1 - b);
		else ng = (n + 1 - a) * (n + 1 - b) - (n + 2 - a - b) * (n + 1 - a - b);
		ng %= mod;
		
		int ans = (all - ng * ng % mod + mod) % mod;
		cout << ans << endl;
	}
	return 0;
}