The following system of linear equations  $2 x+3 y+2 z=9$ ; $3 x+2 y+2 z=9$  ;$x-y+4 z=8$

The value of a for which the system of equations ${a^3}x + {(a + 1)^3}y + {(a + 2)^3}z = 0,$ $ax + (a + 1)y + (a + 2)z = 0,$ $x + y + z = 0,$ has a non zero solution is

$2x + 3y + 4z = 9$,$4x + 9y + 3z = 10,$$5x + 10y + 5z = 11$ then the value of $ x$ is

If the system of equations

$x+y+z=2$

$2 x+4 y-z=6$

$3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then 

If ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}x&3\\{ - 2}&1\end{array}\,} \right|$, then $ x =$

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