Explanation:

In order to understand why a relation scheme in BCNF is also in first normal form (1NF), second normal form (2NF), and third normal form (3NF), it is important to understand the definitions of these normal forms.

1. First Normal Form (1NF):
A relation scheme is said to be in 1NF if and only if all the attributes in the relation are atomic, i.e., they cannot be further divided.

2. Second Normal Form (2NF):
A relation scheme is said to be in 2NF if and only if it is in 1NF and every non-key attribute is fully functionally dependent on the entire primary key.

3. Third Normal Form (3NF):
A relation scheme is said to be in 3NF if and only if it is in 2NF and there are no transitive dependencies, i.e., no non-key attribute is transitively dependent on the primary key.

BCNF (Boyce-Codd Normal Form):
A relation scheme is said to be in BCNF if and only if for every non-trivial functional dependency X → Y, X is a superkey. In other words, in BCNF, every determinant (X) in a functional dependency is a candidate key.

Explanation of the answer:

When a relation scheme is in BCNF, it means that every determinant in a functional dependency is a candidate key. This implies that the relation scheme is also in 1NF because the attributes are required to be atomic (indivisible) in order to form candidate keys.

Additionally, since every determinant is a candidate key, there can be no partial dependencies, which are the primary cause of violations of 2NF. Therefore, a relation scheme in BCNF is also in 2NF.

Furthermore, since there are no partial dependencies in a relation scheme in BCNF, there can also be no transitive dependencies, which are the primary cause of violations of 3NF. Therefore, a relation scheme in BCNF is also in 3NF.

Hence, a relation scheme in BCNF satisfies the requirements of 1NF, 2NF, and 3NF, making the correct answer option 'D' - All of the above.