If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$,then $x =$

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

Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,where $0 \le \theta < 2\pi$,then:

The values of $x$ in the following determinant equation,$\left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$ are

The area of the triangle formed by the points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$ is:

The value of the determinant $ \left|\begin{array}{ccc}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| $ is