$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$

If $a, b, c$ are real,then the value of the determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right| = 1$ if

Let $\theta \in \left(0, \frac{\pi}{2}\right)$. If the system of linear equations
$(1+\cos^2 \theta) x + \sin^2 \theta y + 4 \sin 3\theta z = 0$
$\cos^2 \theta x + (1+\sin^2 \theta) y + 4 \sin 3\theta z = 0$
$\cos^2 \theta x + \sin^2 \theta y + (1+4 \sin 3\theta) z = 0$
has a non-trivial solution,then the value of $\theta$ is:

The value of $\left|\begin{array}{lll}x & p & q \\ p & x & q \\ p & q & x\end{array}\right|$ is

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

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval