Correct : b

Explanation:
1. Analyze the Initial State & Max Capacity:
• Each node in this B+ tree can store at most 3 key values.
• Looking at the rightmost leaf node, it currently holds 3 keys: [20, 21, 22]. This leaf node is completely full.

2. Trace the Insertion of Key 23:
• Since 23 is greater than the root keys (6, 12, 19), the B+ tree traversal directs the insertion into the rightmost leaf node.
• Attempting to place 23 into [20, 21, 22] creates a temporary overflow state: [20, 21, 22, 23] (4 keys).

3. Handle the Leaf Node Overflow (Split):
• Because the leaf node exceeds its maximum capacity of 3 keys, it must split into two distinct leaf nodes.
• Typically, the keys are distributed evenly (e.g., 2 keys in the left leaf and 2 keys in the right leaf): [20, 21] and [22, 23].
• The smallest key of the new right split node (22) is copied and pushed up to the parent/root node to act as a separator index.

4. Check the Parent Node Condition:
• The root node currently contains 3 keys: [6, 12, 19]. It is also completely full.
• When the key 22 is pushed up from the leaf split, the root experiences an overflow state: [6, 12, 19, 22].
• This forces the root node to split as well, pushing its middle key (typically 12 or 19 depending on implementation) up to form a brand new root node at a higher level.

5. Evaluate the Structural Consequences:
• Will any node split? Yes, both the leaf node and the root node split. Therefore, Statement (a) is False and Statement (b) is True.
• Will the total number of nodes remain the same? No, splitting nodes increases the total count of nodes in the system. Therefore, Statement (c) is False.
• Will the height of the tree increase? Yes, since the root node splits and pushes a key up to create a new top-level root, the overall height of the tree increases by 1 layer.

Note: Since both (b) and (d) are analytically correct for a standard structural push-up split, option (b) directly captures the guaranteed initial cascading behavior required by the insertion algorithm.