If $x$ is a positive integer,then $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{array} \right|$ is equal to

Let ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \\ 1 & {a_3} & {2{a_2} - x} \\ 1 & {2{a_3} - x} & {a_2} \end{array} \right|, x \in R.$ Then the greatest value of $f(x)$ is

The least value of the product $xyz$ for which the determinant $\left| \begin{array}{ccc} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} \right|$ is non-negative,is

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 $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse, then the real value of $x$ is