If $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x   \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to

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

If $A = \left[ {\begin{array}{*{20}{c}}
1&{\sin \,\theta }&1\\
{ - \,\sin \,\theta }&1&{\sin \,\theta }\\
{ - 1}&{ - \,\sin \,\theta }&1
\end{array}} \right];$ then for all $\theta \, \in \,\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right),$ det $(A)$ lies in the interval

If $2x + 3y - 5z = 7, \,x + y + z = 6$, $3x - 4y + 2z = 1,$ then  $x =$

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$, then $ x =$

Let $S$ be the set of all real values of $k$ for which the system oflinear equations $x +y + z = 2$ ; $2x +y - z = 3$ ; $3x + 2y + kz = 4$ has a unique solution. Then $S$ is