If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :

Evaluate the determinants : $\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$

$\left| {\,\begin{array}{*{20}{c}}5&3&{ - 1}\\{ - 7}&x&{ - 3}\\9&6&{ - 2}\end{array}\,} \right| = 0$, then $ x$ is equal to

If $A, B, C$  be the angles of a triangle, then $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}\,} \right| = $

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$  has the value

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is