Let M=beginpmatrix1&k&0\\0&1&k\k&0&1endpmatrix where k is a real number.

Expanding along the first row gives $\det M=1+k^3=(k+1)(k^2-k+1)$, and the quadratic factor has negative discriminant, so $M$ is singular only at $k=-1$. At $k=-1$ the rows are $(1,-1,0),(0,1,-1),(-1,0,1)$, which sum to zero, while the first two are independent, so the rank is $2$: (A) is true. For $k=2$, $\det M=9$ and $\det(\operatorname{adj}M)=(\det M)^{n-1}=9^2=81$, not $27$: (B) is false. For $k=1$, $\det M=2$ and $M^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$ with trace $\frac32$, not $2$: (C) is false. Finally $M$ is singular for exactly one real $k$, not two: (D) is false.