The system of linear equations $x + \lambda y - z = 0,\lambda x - y - z = 0\;,\;x + y - \lambda z = 0$ has a non-trivial solution for:

Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero

Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :

If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, then $k =$

The values of $x $ in the following determinant equation, $\left| {\,\begin{array}{*{20}{c}}{a + x}&{a - x}&{a - x}\\{a - x}&{a + x}&{a - x}\\{a - x}&{a - x}&{a + x}\end{array}\,} \right| = 0$ are

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

If the system of linear equations $x + y + z = 5$ ; $x = 2y + 2z = 6$ ; $x + 3y + \lambda z = u (\lambda \, \mu \in R)$, has infinitely many solutions then the value of $\lambda  + \mu $ is