For p > 0 , a vector v → 2 = 2 i ^ + p + 1 j ^ is obtained by rotating the vector v → 1 = 3…

We have,

${\overrightarrow{v}}_1=\sqrt{3}p\hat{i}+\hat{j}$

${\overrightarrow{v}}_2=2\hat{i}+\left(p+1\right)\hat{j}$

$\tan \theta =\frac{\left(\alpha \sqrt{3}-2\right)}{\left(4\sqrt{3}+3\right)}$

$\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|$

$\Rightarrow 3p^2+1=4+(p+1{)}^2$

$\Rightarrow 2p^2-2p-4=0$

$\Rightarrow p^2-p-2=0$

$\Rightarrow p=2, -1$ (rejected)

Now,

$\cos \theta =\frac{{\overrightarrow{v}}_1·{\overrightarrow{v}}_2}{\left|{\overrightarrow{v}}_1\right|·\left|{\overrightarrow{v}}_2\right|}=\frac{2\sqrt{3}p+(p+1)}{\sqrt{(p+1{)}^2+4}\sqrt{3p^2+1}}$

$\Rightarrow \cos \theta =\frac{4\sqrt{3}+3}{\sqrt{13}\sqrt{13}}=\frac{4\sqrt{3}+3}{13}$

$\Rightarrow \tan \theta =\frac{\sqrt{112-24\sqrt{3}}}{4\sqrt{3}+3}=\frac{\sqrt{{\left(6\sqrt{3}-2\right)}^2}}{4\sqrt{3}+3}$

$\Rightarrow \tan \theta =\frac{6\sqrt{3}-2}{4\sqrt{3}+3}=\frac{\alpha \sqrt{3}-2}{4\sqrt{3}+3}$

$\Rightarrow \alpha =6$