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

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

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is:

The area of a triangle is $5$. If two of its vertices are $(2, 1)$ and $(3, -2)$ and the third vertex lies on the line $y = x + 3$,then the third vertex is

The values of $x$ for which the given matrix $\left[\begin{array}{ccc}-x & x & 2 \\ 2 & x & -x \\ x & -2 & -2\end{array}\right]$ will be non-singular are

The solutions of $\operatorname{det}(A-\lambda I_2)=0$ are $4$ and $8$, where $A=\begin{bmatrix} 2 & 3 \\ x & y \end{bmatrix}$. Then: