If $x = a + 2b$ satisfies the cubic $(a, b\in R)$ $f (x)=$ $\left| {\,\begin{array}{*{20}{c}}{a - x}&b&b\\b&{a - x}&b\\b&b&{a - x}\end{array}\,} \right|$ $= 0$, then its other two roots are

$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&c\\b&{ - c}&0\end{array}} \right| = $

Let $a_1,a_2,a_3,....,a_{10}$ be in $G.P.$ with $a_i > 0$ for $i = 1, 2,....,10$ and $S$ be the set of pairs $(r,k), r, k \in N$ (the set of natural numbers) for which

$\left| {\begin{array}{*{20}{c}}
  {{{\log }_e}\,a_1^ra_2^k}&{{{\log }_e}\,a_2^ra_3^k}&{{{\log }_e}\,a_3^ra_4^k} \\
  {{{\log }_e}\,a_4^ra_5^k}&{{{\log }_e}\,a_5^ra_6^k}&{{{\log }_e}\,a_6^ra_7^k} \\ 
  {{{\log }_e}\,a_7^ra_8^k}&{{{\log }_e}\,a_8^ra_9^k}&{{{\log }_e}\,a_9^ra_{10}^k} 
\end{array}} \right| = 0$

Then the number of elements in $S$, is

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ are

The following system of linear equations  $7 x+6 y-2 z=0$ ; $3 x+4 y+2 z=0$ ; ${x}-2{y}-6{z}=0,$ has

An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x  + \beta y + 2z = 2$ has a unique solution, is