Due to lanthanoid contraction, the size of $Ln^{3+}$ ions decreases regularly from $La^{3+}$ to $Lu^{3+}$. As the size of the cation decreases, the covalent character of the $Ln-OH$ bond increases (according to Fajan's rules). Consequently, the basic strength of the hydroxides steadily decreases from $La(OH)_3$ to $Lu(OH)_3$. Thus, $La(OH)_3$ is the most basic and $Lu(OH)_3$ is the least basic. Therefore, the statement '$La(OH)_3$ is less basic than $Lu(OH)_3$' is incorrect. The other statements are correct: the ionic radius of $Ln^{3+}$ ions decreases along the series , La ($Z=57$) is considered a d-block transition element , and the atomic radii of Zr (160 pm) and Hf (159 pm) are almost identical due to lanthanoid contraction .