For any $a, b, c \in R$,the determinant $\left|\begin{array}{lll}bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1\end{array}\right|$ is equal to

The area of the triangle with vertices $(a, b)$,$(x_1, y_1)$,and $(x_2, y_2)$,where $a, x_1, x_2$ are in $G.P.$ with common ratio $r$ and $b, y_1, y_2$ are in $G.P.$ with common ratio $s$,is given by

The number of distinct real roots of the equation $\begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x \end{vmatrix} = 0$ in the interval $\left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$ is

If $\left| \begin{matrix} 1 & a & a^2 \\ 1 & x & x^2 \\ b^2 & ab & a^2 \end{matrix} \right| = 0$,then:

Let $f(\theta) = \left| \begin{array}{ccc} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{array} \right|$. Suppose $A$ and $B$ are respectively the maximum and minimum values of $f(\theta)$. Then $(A, B)$ 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: