$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $

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 $\mathrm{k} \neq 3$.

If the system of equations

$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \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, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

The number of real values of $\lambda $ for which the system of linear equations $2x + 4y - \lambda  z = 0$ ;$4x + \lambda y + 2z = 0$ ; $\lambda x + 2y+ 2z = 0$ has infinitely many solutions, is

Let $a ,b ,c $ be such that $b + c \ne 0$  if

$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to