Given $$|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$$

$$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$$

Dot product with $$\overrightarrow{\mathrm{a}}$$ on both sides

$$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$$ ..... (1)

Dot product with $$\vec{b}$$ on both sides

$$\begin{aligned} & \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\ & |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\ & |\overrightarrow{\mathrm{c}}|^2=48+144 \\ & |\overrightarrow{\mathrm{c}}|^2=192 \\ & \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\ & \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\ & \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\ & \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\ & \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \end{aligned}$$