Evaluate the determinants
$\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$
$-10$
$-11$
$-12$
$-13$
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
The system of equations : $2x\, \cos^2\theta + y\, \sin2\theta - 2\sin\theta = 0$ $x\, \sin2\theta + 2y\, \sin^2\theta = - 2\, \cos\theta$ $x\, \sin\theta - y \cos\theta = 0$ , for all values of $\theta$ , can
If $\left| {{\kern 1pt} \begin{array}{*{20}{c}}1&2&3\\2&x&3\\3&4&5\end{array}\,} \right| = 0,$ then $x =$
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
If $\mathrm{a}_{\mathrm{r}}=\cos \frac{2 \mathrm{r} \pi}{9}+i \sin \frac{2 \mathrm{r} \pi}{9}, \mathrm{r}=1,2,3, \ldots, i=\sqrt{-1}$ then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to :