If $\left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \\ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}\right| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$,then $a+b+c+d+e+f$ is equal to

Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$

The system of linear equations $x + \lambda y - z = 0, \lambda x - y - z = 0, x + y - \lambda z = 0$ has a non-trivial solution for:

The equation obtained by eliminating $a, b, c$ from the equations $x = \frac{a}{b-c}$,$y = \frac{b}{c-a}$,and $z = \frac{c}{a-b}$ is

Let $\sigma_1, \sigma_2, \sigma_3$ be planes passing through the origin. Assume that $\sigma_1$ is perpendicular to the vector $(1, 1, 1)$,$\sigma_2$ is perpendicular to a vector $(a, b, c)$,and $\sigma_3$ is perpendicular to the vector $(a^2, b^2, c^2)$. What are all the positive values of $a, b$,and $c$ so that $\sigma_1 \cap \sigma_2 \cap \sigma_3$ is a single point?

If $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0$,then the area of the triangle whose vertices are $\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), \left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$ is: