In solving a system of linear equations $AX=B$ by Cramer's rule,in the usual notation,if $\Delta_1=\left|\begin{array}{ccc}-11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ccc}4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5\end{array}\right|$,then $X=$

Let $\lambda$ be a real number for which the system of linear equations $x + y + z = 6$,$4x + \lambda y - \lambda z = \lambda - 2$,and $3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:

Let $\beta$ be a real number. Consider the matrix $A = \begin{bmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{bmatrix}$. If $A^7 - (\beta - 1)A^6 - \beta A^5$ is a singular matrix,then the value of $9\beta$ is:

If the system of equations $3x - 2y + z = 0$, $\lambda x - 14y + 15z = 0$, and $x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$, then $\lambda = $

If the system of equations $2x + 3y - z = 5$,$x + \alpha y + 3z = -4$,and $3x - y + \beta z = 7$ has infinitely many solutions,then $13\alpha\beta$ is equal to

Statement $-1$: The system of linear equations
$x + (\sin \alpha)y + (\cos \alpha)z = 0$
$x + (\cos \alpha)y + (\sin \alpha)z = 0$
$x - (\sin \alpha)y - (\cos \alpha)z = 0$
has a non-trivial solution for only one value of $\alpha$ lying in the interval $(0, \frac{\pi}{2})$.
Statement $-2$: The equation in $\alpha$
$\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right| = 0$
has only one solution lying in the interval $(0, \frac{\pi}{2})$.