If $A$, $B$ and $C$ are square matrices of order $3$ such that $A = \left[ {\begin{array}{*{20}{c}}   x&0&1 \\    0&y&0 \\    0&0&z  \end{array}} \right]$ and $\left| B \right| = 36$, $\left| C \right| = 4$,  $\left( {x,y,z \in N} \right)$ and $\left| {ABC} \right| = 1152$ then the minimum value of $x + y + z$ is

Statement $-1$ : The system of linear equations

$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$

$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$

$x – \left( {\sin \,\alpha } \right)y – \left( {\cos \alpha } \right)z = 0$

has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$ 

Statement $-2$ : The equation in $\alpha $

$\left| {\begin{array}{*{20}{c}}
  {\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\ 
  {\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\ 
  {\cos {\mkern 1mu} \alpha }&{ – \sin {\mkern 1mu} \alpha }&{ – \cos {\mkern 1mu} \alpha } 
\end{array}} \right| = 0$

has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

Evaluate the determinants

$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval

$\left| {\,\begin{array}{*{20}{c}}5&3&{ – 1}\\{ – 7}&x&{ – 3}\\9&6&{ – 2}\end{array}\,} \right| = 0$, then $ x$ is equal to