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 

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is

For non zero, $a,b,c$ if $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = 0$, then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

$\left| {\,\begin{array}{*{20}{c}}{1/a}&1&{bc}\\{1/b}&1&{ca}\\{1/c}&1&{ab}\end{array}\,} \right| = $

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

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ are