The set of all values of $\lambda $ for which the system of linear equations $x - 2y - 2z = \lambda x$ ; $x + 2y + z = \lambda y$ ; $-x - y = \lambda z$ has non zero solutions.

If a system of the equation ${(\alpha + 1)^3}x + {(\alpha + 2)^3}y - {(\alpha + 3)^3} = 0$ and $(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0,x + y - 1 = 0$ is constant. what is the value of $\alpha $

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

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 $\left( {0\,,\,\frac{\pi }{2}} \right)$

Which of the following is correct?