If the elements of matrix $A$ are the reciprocals of elements of matrix $\left[\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right]$,where $\omega$ is a complex cube root of unity,then:

Among the statements:
$I$: If $\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$, then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$
$II$: If $\begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$, then $p^{2}=196q^{2}$

If $A = \begin{bmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{bmatrix}$ and $\det\left(A^{2} - \frac{1}{2} I\right) = 0$,then a possible value of $\alpha$ is

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 $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$,$A^2 = A^T$,then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to . . . . . . .

The value of $\theta$ lying between $-\frac{\pi}{4}$ and $\frac{\pi}{2}$ and $0 \le A \le \frac{\pi}{2}$ satisfying the equation $\begin{vmatrix} 1 + \sin^2 A & \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & 1 + \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & \cos^2 A & 1 + 2 \sin 4\theta \end{vmatrix} = 0$ are: