The system of equations $x - 2y + 3z = 5$,$2x - 2y + z = 0$,and $-x + 2y - 3z = 6$ has

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$, $5x+2y-4z+9=0$ and $8x-6y-z+5=0$, then

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 $(0, \frac{\pi}{2})$.

Let $A = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix}$, where $a$ and $b$ are the roots of the equation $x^2 - 4x + 2 = 0$. If $A + A^{-1} = kI_2$, then the value of $k$ is . . . . . .

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

Let $A = \begin{bmatrix} 12 & 24 & 5 \\ x & 6 & 2 \\ -1 & -2 & 3 \end{bmatrix}$. The value of $x$ for which the matrix $A$ is not invertible is