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

Let $d \in R$, and  $A = \left[ {\begin{array}{*{20}{c}} { – 2}&{4 + d}&{\left( {\sin \,\theta } \right) – 2}\\ 1&{\left( {\sin \,\theta } \right) + 2}&d\\ 5&{\left( {2\sin \,\theta } \right) – d}&{\left( { – \sin \,\theta } \right) + 2 + 2d} \end{array}} \right]$, $\theta  \in \left[ {0,2\pi } \right]$. If the minimum value of det $(A)$ is $8$, then a value of $d$ is

$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $

If the system of equations $\mathrm{x}+4 \mathrm{y}-\mathrm{z}=\lambda$, $7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu .+3 \lambda)$ is equal to :

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$