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

If ${A_\lambda } = \left( {\begin{array}{*{20}{c}}
\lambda &{\lambda  - 1}\\
{\lambda  - 1}&\lambda 
\end{array}} \right);\,\lambda  \in N$ then $|A_1| + |A_2| + ..... + |A_{300}|$ is equal to

The system of equations $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $, has no solution for

If the system of linear equations $x+y+3 z=0$

$x+3 y+k^{2} z=0$

$3 x+y+3 z=0$

has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to

$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $

Let $\mathrm{A}(-1,1)$ and $\mathrm{B}(2,3)$ be two points and $\mathrm{P}$ be a variable point above the line $A B$ such that the area of $\triangle \mathrm{PAB}$ is $10$ . If the locus of $\mathrm{P}$ is $\mathrm{ax}+\mathrm{by}=15$, then $5 a+2 b$ is :