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 $a \ne 6,b,c$ satisfy $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$then $abc = $

If ${D_p} = \left| {\,\begin{array}{*{20}{c}}p&{15}&8\\{{p^2}}&{35}&9\\{{p^3}}&{25}&{10}\end{array}\,} \right|$, then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = $

Let $a,b,c$ be positive real numbers. The following system of equations in $x, y$  and $ z $ $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} – \frac{{{z^2}}}{{{c^2}}} = 1$, $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1, – \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$ has