If points $(5, 5)$,$(10, k)$ and $(-5, 1)$ are collinear,then $k =$

If $\left| \begin{array}{ccc} \cos 2x & \sin^2 x & \cos 4x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 4x & \cos^2 x & \cos 2x \end{array} \right| = a_0 + a_1 \sin x + a_2 \sin^2 x + \dots$,then $a_0$ is equal to:

Let $\alpha, \beta, \gamma$ be the real roots of the equation $x^{3} + ax^{2} + bx + c = 0$,where $a, b, c \in R$ and $a, b \neq 0$. If the system of equations in $u, v, w$ given by $\alpha u + \beta v + \gamma w = 0$,$\beta u + \gamma v + \alpha w = 0$,and $\gamma u + \alpha v + \beta w = 0$ has a non-trivial solution,then the value of $\frac{a^{2}}{b}$ is:

The area of a triangle with vertices $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$ is $6$ sq. units. Find the value of $k$.

If $\left| \begin{array}{ccc} x + 1 & x + 2 & x + 3 \\ x + 2 & x + 3 & x + 4 \\ x + a & x + b & x + c \end{array} \right| = 0$,then $a, b, c$ are in

If $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} > 0$,then $abc >$