Step 1: Translate each commutator into a physical statement.
A zero commutator means the two observables are compatible, they can be measured together with no fundamental limit, and their operators share eigenvectors. A non-zero commutator means the observables are incompatible, no state can be an exact eigenstate of both at once.
Step 2: Read off compatibility for each pair.
$\hat{A}$ and $\hat{B}$ are compatible ($[\hat{A},\hat{B}]=0$). $\hat{B}$ and $\hat{C}$ are compatible ($[\hat{B},\hat{C}]=0$). $\hat{A}$ and $\hat{C}$ are incompatible ($[\hat{A},\hat{C}] \neq 0$).
Step 3: Check statement (C) using compatibility.
Compatible observables always admit a shared eigenbasis, that is the entire meaning of simultaneous diagonalization. Since $\hat{A}$ and $\hat{B}$ are compatible, this shared basis exists, so (C) is TRUE.
Step 4: Check statement (D) using compatibility.
A single basis that diagonalizes all of $\hat{A}, \hat{B}, \hat{C}$ together would in particular be a shared eigenbasis for $\hat{A}$ and $\hat{C}$ too, forcing $[\hat{A},\hat{C}]=0$. Since we are told the opposite, no such triple-diagonalizing basis can exist, so (D) is FALSE.
Step 5: Check statement (A) using compatibility.
$\hat{A}$ and $\hat{B}$ are compatible, meaning the system can sit in a state that is a simultaneous eigenstate of both, where each uncertainty is individually zero. That makes $\Delta A \, \Delta B$ equal to zero in such a state, so it is not always greater than zero, and (A) is FALSE.
Step 6: Check statement (B) using compatibility.
$\hat{A}$ and $\hat{C}$ are incompatible. Because no state exists in which both are simultaneously sharp, the system cannot reach $\Delta A = 0$ and $\Delta C = 0$ together, so their product $\Delta A \, \Delta C$ stays above zero. Statement (B) is TRUE.
Final Answer:
Reading each commutator as a compatible or incompatible pair of observables again shows (B) and (C) are the two statements that must hold.
\[ \boxed{\text{B, C}} \]