$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $

Consider the system of linear equations

$-x+y+2 z=0$

$3 x-a y+5 z=1$

$2 x-2 y-a z=7$

Let $S_{1}$ be the set of all $\mathrm{a} \in {R}$ for which the system is inconsistent and $S_{2}$ be the set of all $a \in {R}$ for which the system has infinitely many solutions. If $n\left(S_{1}\right)$ and $n\left(S_{2}\right)$ denote the number of elements in $S_{1}$ and $\mathrm{S}_{2}$ respectively, then

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

If $A = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,B = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,C = \left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,$ then which relation is correct

Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then

The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third  vertex which lies on line $y = x + 3$ is-