The sum of all the roots of the equation

$\left|\begin{array}{ccc}x & -3 & 2 \\ -1 & -2 & (x-1) \\ 1 & (x-2) & 3\end{array}\right|=0$ is

If $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|=A \lambda^3+B \lambda^2+C \lambda+D$, then $D+A=$

If $A+2 B=\left[\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right]$ and $2 A-B=\left[\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]$, then $\operatorname{tr}(A)-\operatorname{tr}(B)=$

$A, C$ are $3 \times 3$ matrices $B, D$ are $3 \times 1$ matrices. If $A X=B$ has unique solution and $C X=D$ has infinite number of solutions, then