= Solution
Use the full $3\times3$ <matrix> $B$ in the original PDF; the transcription has dropped its middle column. For either <matrix>, the three-dimensional <characteristic polynomial> is $\chi_M(t)=t^3-(\operatorname{tr}M)t^2+[(\operatorname{tr}M)^2-\operatorname{tr}(M^2)]t/2-\det M$. Both have zero <trace> and zero <determinant>, while $\operatorname{tr}(B^2)=0$ and $\operatorname{tr}(C^2)=72$. Consequently
$$
\boxed{\chi_B(t)=t^3,\qquad\chi_C(t)=t(t-6)(t+6).}
$$
For $B$, a useful check is
$$
B^2=\begin{pmatrix}12&0&-24\\18&0&-36\\6&0&-12\end{pmatrix},\qquad B^3=0,
$$
for every value of $\alpha$. Thus $B$ is a nonzero <nilpotent matrix> for all $\alpha$; for example its second column never vanishes. By the preceding result, $\boxed{B\text{ is not diagonalizable for any }\alpha}$.
For $C$, the three <eigenvalues> $-6,0,6$ are distinct and real, so their <eigenvectors> are linearly independent. An explicit eigenbasis is
$$
v_- =\begin{pmatrix}-1\\0\\1\end{pmatrix},\qquad v_0=\begin{pmatrix}2\\-3\\1\end{pmatrix},\qquad v_+=\begin{pmatrix}2\\3\\1\end{pmatrix}.
$$
With $P=(v_-,v_0,v_+)$, $\det P=18\neq0$ and $P^{-1}CP=\operatorname{diag}(-6,0,6)$. Hence $\boxed{C\text{ is diagonalizable over }\mathbb R\text{ and }\mathbb C}$.
A repeated root of the <characteristic polynomial> alone would not settle diagonalizability; for $B$ it is the nonzero nilpotence that completes the argument.
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