Let the system of equations $x+2 y+3 z=5$, $2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :

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

if $\left| \begin{gathered}
   – 6\ \ \,\,1\ \ \,\,\lambda \ \  \hfill \\
  \,0\ \ \,\,\,\,3\ \ \,\,7\ \  \hfill \\
   – 1\ \ \,\,0\ \ \,\,5\ \  \hfill \\ 
\end{gathered}  \right| = 5948 $, then $\lambda $  is

If the system of linear equations  $x-2 y+z=-4 $   ;  $2 x+\alpha y+3 z=5 $  ;  $3 x-y+\beta z=3$ has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to

The existence of unique solution of the system of equations,  $x+y+z=\beta $ , $5x-y+\alpha z=10$ , $2x+3y-z=6$ depends on