Find the area of the triangle whose vertices are $(3,8),(-4,2)$ and $(5,1)$

If the system of linear equations $x + y + z = 5$ ; $x = 2y + 2z = 6$ ; $x + 3y + \lambda z = u (\lambda \, \mu \in R)$, has infinitely many solutions then the value of $\lambda  + \mu $ is

If $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x   \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to

If $\omega $ be a complex cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}1&\omega &{ - {\omega ^2}/2}\\1&1&1\\1&{ - 1}&0\end{array}\,} \right| = $

$S$ denote the set of all real values of $\lambda$ such that the system of equations  $\lambda x + y + z =1$ ; $x +\lambda y + z =1$ ; $x + y +\lambda z =1$ is inconsistent, then $\sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)$ is equal to

The system of equations $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $, has no solution for