If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$, then $f(100)$ is equal to:

Evaluate the determinant: $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

If $\Delta = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$,then $\Delta$ lies in the interval

$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .

Evaluate the determinant: $\left| \begin{array}{ccc} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{array} \right|$

If $1$,$\log_{10}(4^{x}-2)$ and $\log_{10}(4^{x}+\frac{18}{5})$ are in arithmetic progression for a real number $x$,then the value of the determinant $\left|\begin{array}{ccc} 2(x-\frac{1}{2}) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{array}\right|$ is equal to ...... .