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

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

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

Find area of the triangle with vertices at the point given in each of the following: $(2,7),(1,1),(10,8)$

If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
  {x + 2}&\omega &{{\omega ^2}} \\ 
  \omega &{x + 1 + {\omega ^2}}&1 \\ 
  {{\omega ^2}}&1&{x + 1 + \omega } 
\end{array}} \right| = 0$ is 

The maximum value of

$f(x)=\left|\begin{array}{ccc} \sin ^{2} x & 1+\cos ^{2} x & \cos 2 x \\ 1+\sin ^{2} x & \cos ^{2} x & \cos 2 x \\ \sin ^{2} x & \cos ^{2} x & \sin 2 x \end{array}\right|, x \in R \text { is }$