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

Let $\alpha, \beta \in \mathbb{R}$ be such that the system of linear equations

$ \begin{aligned} x + 2y + z &= 5 \\ 2x + y + \alpha z &= 5 \\ 8x + 4y + \beta z &= 18 \end{aligned} $

has no solution. Then $\frac{\beta}{\alpha}$ is equal to :

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$, then among the two statements:

(S1) : $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$

and

(S2) : $\det(\mathrm{adj}(A+B)) = -5$

Let $A, B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B=(I+A)^{-1}$ and $\mathrm{A}+\mathrm{C}=\mathrm{I}$.

If $\mathrm{BC}=\left[\begin{array}{cc}1 & -5 \\ -1 & 2\end{array}\right]$ and $\mathrm{CB}\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}12 \\ -6\end{array}\right]$, then $x_1+x_2$ is