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Find the value of $x$,if $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{ll}x & 3 \\ 2x & 5\end{array}\right|$.

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

If $a, b, c$ are real numbers such that $a-b=1$ and $b-c=3$,then the number of matrices of the form $A=\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ such that $|A|=-12$ is:

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:

If $\alpha, \beta, \gamma$ $(\alpha < \beta < \gamma)$ are the values of $x$ such that $\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = 0$ is a singular matrix,then $2\alpha + 3\beta + 4\gamma = $

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