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:

The solutions of the equation $\left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$ are:

Let $A = \begin{bmatrix} 3-t & 1 & 0 \\ -1 & 3-t & 1 \\ 0 & -1 & 0 \end{bmatrix}$ and $\det(A) = 5$, then find the value of $t$.

If $\left| \begin{array}{ccc} 1 & 2 & 3 \\ 2 & x & 3 \\ 3 & 4 & 5 \end{array} \right| = 0$,then $x =$

If $A$,$B$ and $C$ are the angles of a triangle,then the value of the determinant $\left| \begin{array}{ccc} -1 + \cos B & \cos C + \cos B & \cos B \\ \cos C + \cos A & -1 + \cos A & \cos A \\ -1 + \cos B & -1 + \cos A & -1 \end{array} \right|$ is:

If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.