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 the numbers $2, b, c$ be in an $A.P.$ and $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}$. If $\det(A) \in [2, 16]$,then $c$ lies in the interval:

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:

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & {\sin \theta } & {\cos \theta } \\ {\sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & {\sin 2\theta } & {\cos 2\theta } \\ {\sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \end{array}} \right|$,$x \ne 0$; then for all $\theta \in \left( {0, \frac{\pi }{2}} \right)$:

In a square matrix $A$ of order $3$,$a_{ii} = i + m_i$,where $i = 1, 2, 3$ and $m_i$ are the slopes (in increasing order of their absolute value) of the $3$ normals concurrent at the point $(9, -6)$ to the parabola $y^2 = 4x$. All other entries of the matrix are $1$. The value of $\det(A)$ is equal to

Let $S$ be the set containing all $3 \times 3$ matrices with entries from $\{-1, 0, 1\}$. The total number of matrices $A \in S$ such that the sum of all the diagonal elements of $A^{T}A$ is $6$ is: