If $\left|\begin{array}{ll}2017 & 2018 \\ 2019 & 2020\end{array}\right|+\left|\begin{array}{ll}2021 & 2022 \\ 2023 & 2024\end{array}\right|=2 k$,then $k^3=$ . . . . . .

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $i = 1, 2, 3$ and $a_ia_j + b_ib_j + c_ic_j = 0$ for $i \ne j$ where $i, j = 1, 2, 3$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$ is:

The area of a triangle is $13$ sq. units whose vertices are $A(8, 2)$,$B(k, 4)$,and $C(6, 7)$. Then,the integer value of $k$ is . . . . . . .

Find the area of the triangle with vertices $(a, b)$,$(x_1, y_1)$,and $(x_2, y_2)$,where $a, x_1, x_2$ are in $G.P.$ with common ratio $r$,and $b, y_1, y_2$ are in $G.P.$ with common ratio $s$.

The value of $\left| \begin{array}{ccc} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{array} \right|$ is equal to

$\left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}} \right| = $