If $t_1, t_2$ and $t_3$ are distinct,then the points $(t_1, 2at_1 + at_1^3)$,$(t_2, 2at_2 + at_2^3)$ and $(t_3, 2at_3 + at_3^3)$ are collinear if:

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

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:

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^{3}| = 125$,then $\alpha$ is equal to

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:

Find the area of the triangle with vertices at the points $(1,0), (6,0), (4,3)$.