The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then

Consider the system of equations

$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$

$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and

$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.

The system of equations $kx + 2y\,-z = 1$  ;  $(k\,-\,1)y\,-2z = 2$  ;  $(k + 2)z = 3$ has unique solution, if $k$ is equal to

Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :

The least value of the product $xyz$ for which the determinant $\left| {\begin{array}{*{20}{c}}
  x&1&1 \\ 
  1&y&1 \\ 
  1&1&z 
\end{array}} \right|$ is non-negative, is 

If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes  $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to