The basic strength `"…................"` as we proceed from`La(OH)_(3)` to `Lu(OH)_(3)` ( decreases or increases or remains constant).

Answer

To determine how the basic strength of hydroxides changes as we proceed from La(OH)₃ (Lanthanum Hydroxide) to Lu(OH)₃ (Lutetium Hydroxide), we can analyze the trend in the lanthanide series.

### Step-by-Step Solution:

1. **Understanding the Series**: The lanthanide series consists of elements from Lanthanum (La) to Lutetium (Lu), which are f-block elements. The atomic numbers range from 57 (La) to 71 (Lu).

2. **Basic Strength of Hydroxides**: Basic strength refers to the ability of a compound to donate hydroxide ions (OH⁻) in solution. For hydroxides of lanthanides, this is influenced by the size of the metal ion and its charge density.

3. **Trends in Ionic Radius**: As we move from La to Lu, the ionic radius of the lanthanide ions decreases. This is due to the increasing effective nuclear charge (Z-effective) as protons are added to the nucleus while the electrons are added to the f-orbitals, which do not shield the nuclear charge effectively.

4. **Effect of Ionic Radius on Basic Strength**: A smaller ionic radius means that the metal ion has a higher charge density. This higher charge density leads to a stronger attraction between the metal ion and the hydroxide ion (OH⁻). Consequently, it becomes more difficult for the hydroxide ion to be released into solution.

5. **Conclusion**: Therefore, as we move from La(OH)₃ to Lu(OH)₃, the basic strength of these hydroxides decreases. This is because the decreasing size of the metal ion results in a decreased tendency to release hydroxide ions.

### Final Answer:
The basic strength decreases as we proceed from La(OH)₃ to Lu(OH)₃.

---