If $\omega $ is a cube root of unity and $\Delta = \left| {\begin{array}{*{20}{c}}1&{2\omega }\\\omega &{{\omega ^2}}\end{array}} \right|$, then ${\Delta ^2}$ is equal to

If $a_i^2 + b_i^2 + c_i^2 = 1,\,i = 1,2,3$ and $a_ia_j + b_ib_j +c_ic_j = 0$ $\left( {i \ne j,i,j = 1,2,3} \right)$ then the value of determinant $\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\ 
  {{b_1}}&{{b_2}}&{{b_3}} \\ 
  {{c_1}}&{{c_2}}&{{c_3}} 
\end{array}} \right|$ is

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, then $k =$

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4,3 x+2 y+5 z=3$ $9 x+4 y+(28+[\lambda]) z=[\lambda]$ has a solution is:

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are  $(\mathrm{k}, 0),(4,0),(0,2)$