If $\alpha$ is a real root of the equation $x^3+6x^2+5x-42=0$,then the determinant of the matrix $\left[\begin{array}{ccc}\alpha-1 & \alpha+1 & \alpha+2 \\ \alpha-2 & \alpha+3 & \alpha-3 \\ \alpha+4 & \alpha-4 & \alpha+5\end{array}\right]$ is

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3 \, \text{sq units}$.

$\left| {\begin{array}{ccc} 19 & 17 & 15 \\ 9 & 8 & 7 \\ 1 & 1 & 1 \end{array}} \right| = $

Let $x, y, z > 0$ be the $2^{nd}, 3^{rd}, 4^{th}$ terms of a $G.P.$ respectively,and $\Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} = (r-1)^2 \left(1 - \frac{1}{r^2}\right)$,where $r$ is the common ratio. Then $k = \dots$

If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is: