The determinant $\left| \begin{array}{ccc} a^2 & a^2 - (b - c)^2 & bc \\ b^2 & b^2 - (c - a)^2 & ca \\ c^2 & c^2 - (a - b)^2 & ab \end{array} \right|$ is divisible by :

If each element of a second order determinant is either zero or one,what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently,each value being assumed with probability $\frac{1}{2}$).

If $f:[0, \pi / 2) \rightarrow R$ is defined as $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$, then the range of $f$ is:

Let $p$ and $p+2$ be prime numbers and let $\Delta=\left|\begin{array}{ccc}p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)!\end{array}\right|$. Then the sum of the maximum values of $\alpha$ and $\beta$,such that $p^{\alpha}$ and $(p+2)^{\beta}$ divide $\Delta$,is $........$

Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$,$A^2 = A^T$,then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to . . . . . . .

Let $a, b, c \in \mathbb{R}$ be all non-zero and satisfy $a^{3}+b^{3}+c^{3}=2$. If the matrix $A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$ satisfies $A^{T} A=I$,then a value of $abc$ can be