If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -
 

The value of the determinant given below $\left| {{\rm{ }}\begin{array}{*{20}{c}}1&2&3\\3&5&7\\8&{14}&{20}\end{array}} \right|$ is

If $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $

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 $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to