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

Let $\omega $ be a complex number such that  $2\omega + 1 = z$ where $z = \sqrt { - 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ - {\omega ^2} - 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations

$a x+2 a y-3 a z=1$

$(2 a+1) x+(2 a+3) y+(a+1) z=2$

$(3 a+5) x+(a+5) y+(a+2) z=3$

has unique solution and infinitely many solutions. Then

Let $ \alpha _1, \alpha _2$ are two values of $\alpha $ for which the system $2 \alpha x + y = 5, x - 6y = \alpha $ and $x + y = 2$ is consistent, then $ |2(\alpha _1 + \alpha _2)| $ is -

Let the system of equations $x+2 y+3 z=5$, $2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :

$\left| {\,\begin{array}{*{20}{c}}x&4&{y + z}\\y&4&{z + x}\\z&4&{x + y}\end{array}\,} \right| = $