Let $A=\left[\begin{array}{ccc}1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7\end{array}\right]$ and $\operatorname{det}(A-\alpha I)=0$, where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$, then the circle $(x-p)^2+(y-2 p)^2=320$, intersects the co-ordinate axes at

Let $\mathrm{S}=\left\{\mathrm{A}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in\{0,1,2,3,4\}\right.$ and $\left.\mathrm{A}^2-4 \mathrm{~A}+3 \mathrm{I}=0\right\}$ be a set of $2 \times 2$ matrices. Then the number of matrices in S , for which the sum of the diagonal elements is equal to 4 , is :

Let $A=\left[\begin{array}{ccc}1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5\end{array}\right]$. Then the sum of all elements of the matrix $\operatorname{adj}\left(\operatorname{adj}\left(2(\operatorname{adj} \mathrm{~A})^{-1}\right)\right)$ is equal to:

If the system of equations

$x + 5y + 6z = 4$

$2x + 3y + 4z = 7$

$x + 6y + az = b$

has infinitely many solutions, then the point $(a, b)$ lies on the line