If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$  has the value

$\Delta = \left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + b + c}\\{3a}&{4a + 3b}&{5a + 4b + 3c}\\{6a}&{9a + 6b}&{11a + 9b + 6c}\end{array}\,} \right|$where $a = i,b = \omega ,c = {\omega ^2}$, then $\Delta $is equal to

$\left| {\,\begin{array}{*{20}{c}}0&{p - q}&{p - r}\\{q - p}&0&{q - r}\\{r - p}&{r - q}&0\end{array}\,} \right| = $

$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ - 6}&{ - 2}\\
{ - 6}&{9 + {x^2}}&3\\
{ - 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by

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 = $