A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 - x}&{ - 6}&3\\{ - 6}&{3 - x}&3\\3&3&{ - 6 - x}\end{array}\,} \right| = 0$ is

Let $A=\left(\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right),$ where $[t]$ denotes the greatest integer less than or equal to $\mathrm{t}$. If $\operatorname{det}(\mathrm{A})=192$, then the set of values of $\mathrm{x}$ is the interval

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

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$

Given the system of equation $a(x + y + z)=x,b(x + y + z) = y, c(x + y + z) = z$ where $a,b,c$  are non-zero real numbers. If the real numbers $x,y,z$ are such that $xyz \neq 0,$ then  $(a + b + c)$ is equal to-

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