Prove that the four points `6 hati - 7hatj, 16hati - 19 hatj - 4hatk, 3hatj - 6hatk and 2hati + 5hatj + 10 hatk` form a tetrahedron in spacel.

Explanation

To prove that the four points \( A(6\hat{i} - 7\hat{j}), B(16\hat{i} - 19\hat{j} - 4\hat{k}), C(3\hat{j} - 6\hat{k}), D(2\hat{i} + 5\hat{j} + 10\hat{k}) \) form a tetrahedron, we need to show that these points are non-coplanar. The points are non-coplanar if the scalar triple product of the vectors formed by these points is not equal to zero.

### Step 1: Define the points
Let:
- \( A = 6\hat{i} - 7\hat{j} \)
- \( B = 16\hat{i} - 19\hat{j} - 4\hat{k} \)
- \( C = 3\hat{j} - 6\hat{k} \)
- \( D = 2\hat{i} + 5\hat{j} + 10\hat{k} \)
...