$\left|\begin{array}{lll}24 & 25 & 26 \\ 25 & 26 & 27 \\ 26 & 27 & 27\end{array}\right|$ is equal to

$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is .........

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

$\begin{aligned} & \text { If }\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}\right|=\Delta \text { and } \\ & \left|\begin{array}{ccc}1 & -4 & 7 \\ -2 & 3 & -5 \\ 3 & x & -3\end{array}\right|=2 \Delta+1, \text { then } x=\end{aligned}$

The solution set of $\left|\begin{array}{ccc}x & 3 & 5 \\ 2 & 6 & 10 \\ 7 & 21 & 35\end{array}\right|=0$ is . . . . . . .

If $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,then for all $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$,$\det(A)$ lies in the interval: