Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

has no solution. Then the set $S$

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are $(-2,0),(0,4),(0, \mathrm{k})$

If $n$  be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2\\
2&x&{ - x}\\
x&{ - 2}&{ - x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is

$($ where $det(B)$ denotes determinant of Matrix $B) -$

If ${A_i} = \left[ {\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}} \right]$and if $|a|\, < 1,\,|b|\, < 1$, then $\sum\limits_{i = 1}^\infty {\det ({A_i})} $is equal to

The value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

If the system of equations $x+y+z=6 \,; \,2 x+5 y+\alpha z=\beta  \,; \, x+2 y+3 z=14$ has infinitely many solutions, then $\alpha+\beta$ is equal to.