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

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 value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$

For $0 < \theta < \frac{\pi}{2}$,if $A = \begin{bmatrix} 1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1 \end{bmatrix}$,then which of the following is true regarding $\operatorname{det}(A)$?

If $a$ and $b$ are any two real numbers, then $\left|\begin{array}{ccc} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{array}\right| = $

If $\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right| = $ . . . . . . .