First and foremost, it can be proven that $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$ is always even, for all possible non-negative values of $$$a$$$, $$$b$$$ and $$$c$$$.

Therefore, if $$$n$$$ is even, one possible solution is $$$a=0$$$, $$$b=0$$$ and $$$c=\frac{n}{2}$$$. In this case, $$$(a \oplus b) + (b \oplus c) + (a \oplus c)= 0+\frac{n}{2}+\frac{n}{2}=n$$$. Otherwise, there are no solutions.

Time complexity per testcase: $$$O(1)$$$.

#include<bits/stdc++.h>

using namespace std;

void testcase(){
    int n;
    cin>>n;

    if(n%2==0)
        cout<<"0 "<<n/2<<' '<<n/2<<'\n';
    else
        cout<<"-1\n";
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t;
    cin>>t;
    while(t--)
        testcase();
    return 0;
}

This is actually the third iteration of the problem, which was suggested by antontrygubO_o.

The first iteration had $$$|a-b|+|b-c|+|a-c|=n$$$, and the second one had $$$gcd(a,b)+gcd(b,c)+gcd(a,c)=n$$$.

The general construction consists of a $$$2 \times 2$$$ checkerboard with a $$$1$$$-thick border. Here is the intended solution for $$$n=6$$$ and $$$m=8$$$:

https://codeforces.net/predownloaded/75/c4/75c46e19cc3cf6f890139b0e74774c3a6fc387db.png

Time complexity per testcase: $$$O(nm)$$$.

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

void testcase(){
    ll n,m;
    cin>>n>>m;

    for(ll i=1;i<=n;i++){
        for(ll j=1;j<=m;j++){
            cout<<((i%4<=1)!=(j%4<=1))<<" \n"[j==m];
        }
    }
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t;
    cin>>t;
    while(t--)
        testcase();
    return 0;
}