If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=Ax^{3}+Bx^{2}+Cx+D$,then $B+C$ is equal to

Let $S = \left\{ A = \begin{bmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{bmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}$,where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is:

In a square matrix $A$ of order $3$,$a_{ii}$ are the sum of the roots of the equation $x^2 - (a + b)x + ab = 0$; $a_{i, i+1}$ are the product of the roots,$a_{i, i-1}$ are all unity,and the rest of the elements are all zero. The value of the determinant of $A$ is equal to

If $a, b, c$ are positive and not all equal,then the value of the determinant $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$ is

The area of a triangle with vertices $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$ is $6$ sq. units. Find the value of $k$.

If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .