Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, \mathrm{QR}=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$ and $A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]$, where $I$ is the identity matrix of order $3 \times 3$. If $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ is $2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}$, $\mathrm{n} \in\{0,1,2, \ldots, 20\}$, then $\mathrm{m}+\mathrm{n}$ is equal to :

If $\theta \in[-2 \pi, 2 \pi]$, then the number of solutions of $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0$, is equal to: