If $$\vec a =(1+\sin \theta )\hat i+\cos \theta \hat{ j}+\sin2\theta\hat k\\ \vec b =(\sin( \theta +2\pi/3))\hat i+\cos ( \theta +2\pi/3) \hat{ j}+\sin( 2\theta +4\pi/3)\hat k\\ \vec c =(\sin ( \theta -2\pi/3) )\hat i+\cos( \theta -2\pi/3) \hat{ j}+\sin( 2\theta -4\pi/3)\hat k\\$$ Then if the volume of parallelopiped whose coterminus edges are $2\vec b\times \vec c,3\vec c\times \vec a,4\vec a\times \vec b$ is $4.5$, then how many these satisfy in Open interval of First Quadrant?
I tried solving and used $[2\vec b\times \vec c\;3\vec c\times \vec a\;4\vec a\times \vec b]=24[\vec b\;\vec c\; \vec a]^2$ and formed the required determinant one thing I noted was sum of columns is related to cube-roots of unity, but couldn't proceed.
$$\large V=|[2\vec b\times\vec c\quad3\vec c\times\vec a\quad4\vec a\times\vec b]|=\frac92\\\large 24[\vec a\quad\vec b\quad\vec c]^2=\frac92\implies |[\vec a\quad\vec b\quad\vec c]^2|=\frac3{16}$$ So: $$\large\begin{vmatrix}(1+\sin\theta)&\cos\theta&\sin2\theta\\\sin\left(\theta+\frac{2\pi}3\right)&\cos\left(\theta+\frac{2\pi}3\right)&\sin\left(\theta+\frac{4\pi}3\right)\\\sin\left(\theta-\frac{2\pi}3\right)&\cos\left(\theta-\frac{2\pi}3\right)&\sin\left(2\theta-\frac{4\pi}3\right)\\\end{vmatrix}=\pm\frac{\sqrt3}4$$ Applying $R_1\to R_1+R_2+R_3$ After solving $\cos3\theta=\pm\frac12$ So: $$\implies\theta=\frac{\pi}9,\frac{2\pi}9,\frac{4\pi}9$$