#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if (x < y) x = y; }


template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) { }
	ModInt(signed sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
};
typedef ModInt<1000000007> mint;

int main() {
	int T;
	scanf("%d", &T);
	const int N = 101;
	mint dp[N][N];
	rep(x, N) rep(y, N) {
		mint r;
		if (x == 0 || y == 0) {
			r = 1;
		} else {
			rep(j, y)
				r += dp[x - 1][y - 1 - j];
			rep(i, x)
				r += dp[x - 1 - i][y - 1];
		}
		dp[x][y] = r;
	}
	for (int ii = 0; ii < T; ++ ii) {
		int A; int B; int C;
		scanf("%d%d%d", &A, &B, &C);
		mint ans;
		//頂点を使う場合
		rep(a, A)
			ans += dp[a][B] * dp[A - 1 - a][C - 1];
		rep(b, B)
			ans += dp[b][C] * dp[B - 1 - b][A - 1];
		rep(c, C)
			ans += dp[c][A] * dp[C - 1 - c][B - 1];
		//頂点を使わない場合
		//Lemma 中点のみを使う三角形がある
		//	それがない場合、辺を共有する三角形をたどってくと頂点に行き着くので矛盾。
		rep(a, A) rep(b, B) rep(c, C)
			ans += dp[a][B - 1 - b] * dp[b][C - 1 - c] * dp[c][A - 1 - a];
		printf("%d\n", ans.get());
	}
	return 0;
}