$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $

Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order $3$ . If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}(Q)=\frac{k^2}{2}$, then

($A$) $\quad \alpha=0, k=8$

($B$) $4 \alpha-k+8=0$

($C$) $\operatorname{det}(P \operatorname{adj}(Q))=2^9$

($D$) $\operatorname{det}(Q \operatorname{adj}(P))=2^{13}$

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
  1&1&1 \\ 
  2&b&c \\ 
  4&{{b^2}}&{{c^2}} 
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1,0,1\}$. Then, the maximum possible value of the determinant of $P$ is. . . . . . .

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$