If $A$ is a square matrix of order $3$ and $A^2+A+2I=0$,then

If $A = \begin{bmatrix} p & q & r \\ r & p & q \\ q & r & p \end{bmatrix}$ and $A A^T = I$,then $p^3 + q^3 + r^3 =$ . . . . . .

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 . . . . . . .

Match the statements given in Column $I$ with the intervals/union of intervals given in Column $II$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): |z|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

If $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$,then $x=$

Let $A, B, C, D$ be square real matrices such that $C^T = DAB$,$D^T = ABC$,and $S = ABCD$. Then $S^2$ is equal to: