If $A$ is a square matrix such that $A^2=A$,then $(I-A)^3$ is

If $P$ and $Q$ are square matrices such that $P^{2006} = O$ and $PQ = P + Q$,then $\det(Q)$ will be

Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], \quad P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k \left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$ where $P_k^{\top}$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?
$(1)$ $X - 30I$ is an invertible matrix
$(2)$ The sum of diagonal entries of $X$ is $18$
$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$,then $\alpha=30$
$(4)$ $X$ is a symmetric matrix

Let $\alpha$ be a root of the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$,where $a, b, c$ are distinct real numbers such that the matrix $\begin{bmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{bmatrix}$ is singular. Then the value of $\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$ is:

Let $P$ be a square matrix such that $P^2 = I - P$. For $\alpha, \beta, \gamma, \delta \in N$,if $P^\alpha + P^\beta = \gamma I - 29 P$ and $P^\alpha - P^\beta = \delta I - 13 P$,then $\alpha + \beta + \gamma - \delta$ is equal to $........$.

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] \Rightarrow A^2-2 A=$