Let $|A|=6$ where $A$ is a $3 \times 3$ matrix. If $|adj(3adj(A^{2} \cdot adj(2A)))|=2^{m} \cdot 3^{n}$,$m, n \in N$,then $m+n$ is equal to:

If $A$ and $B$ are square matrices of the same order such that $AB = A$ and $BA = B$,then $(A + I)^5$ is equal to (where $I$ is the identity matrix).

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	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

Let $x \in R$ and let $P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}$,$Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix}$ and $R = PQP^{-1}$. Then which of the following options is/are correct?
$(1)$ For $x = 1$,there exists a unit vector $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ for which $R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
$(2)$ There exists a real number $x$ such that $PQ = QP$.
$(3)$ $\det R = \det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8$,for all $x \in R$.
$(4)$ For $x = 0$,if $R \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix}$,then $a + b = 5$.

The value of $\theta$ lying between $0$ and $\pi / 2$ and satisfying the equation $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ is:

If $A$ and $B$ are $3 \times 3$ order matrices and $|A|=5$,$|B|=3$,then $|3AB|=$ . . . . . . .