The system of equations $kx + y + z =1$ $x + ky + z = k$ and $x + y + zk = k ^{2}$ has no solution if $k$ is equal to

The system of linear equations $x + y + z = 2$, $2x + y – z = 3,$ $3x + 2y + kz = 4$has a unique solution if

If $A_1B_1C_1,\, A_2B_2C_2,\, A_3B_3C_3$ are three digit number each of which is divisible by $k$ and $\Delta  = \left| {\begin{array}{*{20}{c}}
  {{A_1}{\kern 1pt} }&{{B_1}}&{{C_1}} \\ 
  {{A_2}}&{{B_2}}&{{C_2}} \\ 
  {{A_3}}&{{B_3}}&{{C_3}} 
\end{array}} \right|$ ; then $\Delta $ is divisible by

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}0&x&{16}\\x&5&7\\0&9&x\end{array}\,} \right| = 0$  are

If $n \ne 3k$ and 1, $\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^{2n}}}&1&{{\omega ^n}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\end{array}\,} \right|$ has the value