If $-9$ is a root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$,then the other two roots are:

If $\left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right|+\left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right|=\left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2 x & -8 & -11 \\ 5 & 6 & 1\end{array}\right|$,then $x=$ . . . . . .

If ${\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right|^2} = \left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| - \left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right|$,then $x =$

The sum of the real roots of the equation $\left| \begin{matrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{matrix} \right| = 0$ is equal to

Find the area of the triangle with vertices $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$.

If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to