Let $A = \left[ {\begin{array}{*{20}{c}}5&{5\alpha }&\alpha \\0&\alpha &{5\alpha }\\0&0&5\end{array}} \right]$, If ${\left| A \right|^2} = 25$, then $\left| \alpha \right|$ equals

One of the roots of the given equation $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$ is

If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}}
  {{q_1}}&{{q_2}}&{{q_3}} \\ 
  {{q_2}}&{{q_3}}&{{q_1}} \\ 
  {{q_3}}&{{q_1}}&{{q_2}} 
\end{array}} \right|$ is

If $\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2\end{array}\right|=\frac{9}{8}(103 x+81)$, then $\lambda$, $\frac{\lambda}{3}$ are the roots of the equation

The number of $\theta \in(0,4 \pi)$ for which the system of linear equations

$3(\sin 3 \theta) x-y+z=2$, $3(\cos 2 \theta) x+4 y+3 z=3$, $6 x+7 y+7 z=9$ has no solution is.

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $