 Solution

A matrix is diagonalizable if some invertible $S$ makes $S^{-1}AS$ diagonal, equivalently if it has a basis of eigenvectors. If $A^p=0$ and $Av=\lambda v$ for $v\ne 0$, then $0=A^{p}v={\lambda }^{p}v$, so every eigenvalue is zero. There is at least one zero eigenvalue: otherwise $A$ and hence $A^p$ would be invertible. A diagonalizable nilpotent matrix has an all-zero diagonal form and therefore is zero itself.

For the specified matrices, direct determinant expansion givesThe Cayley-Hamilton theorem gives $B^3=0$. Its entry $B_{12}=-2$ is nonzero for every real $\alpha$, so $B$ is not diagonalizable for any $\alpha$. The matrix $C$ has three distinct real eigenvalues and is therefore diagonalizable over $\mathbb{R}$. For an explicit check, take columns $(-1,0,1{)}^T$, $(2,-3,1{)}^T$, $(2,3,1{)}^T$ in $S$; then $S^{-1}CS=\mathrm{diag}(-6,0,6)$.