If the points $(2k, k), (k, 2k)$ and $(k, k)$ with $k > 0$ enclose a triangle of area $18$ square unit then centroid of triangle is equal to

Let $A = \left[ {\begin{array}{*{20}{c}}
  2&b&1 \\ 
  b&{{b^2} + 1}&b \\ 
  1&b&2 
\end{array}} \right]$  where $b > 0$. Then the minimum value of $\frac{{\det \left( A \right)}}{b}$ is

Find area of the triangle with vertices at the point given in each of the following: $(-2,-3),(3,2),(-1,-8)$

Statement $-1$ : The system of linear equations

$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$

$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$

$x – \left( {\sin \,\alpha } \right)y – \left( {\cos \alpha } \right)z = 0$

has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$ 

Statement $-2$ : The equation in $\alpha $

$\left| {\begin{array}{*{20}{c}}
  {\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\ 
  {\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\ 
  {\cos {\mkern 1mu} \alpha }&{ – \sin {\mkern 1mu} \alpha }&{ – \cos {\mkern 1mu} \alpha } 
\end{array}} \right| = 0$

has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then