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

Let $\theta \in\left(0, \frac{\pi}{2}\right)$. If the system of linear equations

$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$

has a non-trivial solution, then the value of $\theta$ is :

The value of $\left| {\begin{array}{*{20}{c}}
1&x&y\\
2&{\sin x + 2x}&{\sin y + 2y}\\
3&{\cos x + 3x}&{\cos y + 3y}
\end{array}} \right|$ is

The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then

One of the roots of the given equation $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$ is

$\left| {\,\begin{array}{*{20}{c}}5&3&{ - 1}\\{ - 7}&x&{ - 3}\\9&6&{ - 2}\end{array}\,} \right| = 0$, then $ x$ is equal to