$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$
Dot product with $\overrightarrow{\mathrm{a}}$ on both sides
$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$
Dot product with $\vec{b}$ on both sides
$ \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 $
$\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 $
$ |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 $
$ |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) $
$ |\overrightarrow{\mathrm{c}}|^2=4[12]+144 $
$ |\overrightarrow{\mathrm{c}}|^2=48+144 $
$ |\overrightarrow{\mathrm{c}}|^2=192 $
$ \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}|} $
$ \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} $
$ \therefore \cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4} $
$ \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} $
$ \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
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