Given that

$\left(x,y\right)=\left(1,2\right)$

$\left(x^{\text{'}},y^{\text{'}}\right)=\left(\frac{3\sqrt{3}-1}{2\sqrt{2}},\frac{\sqrt{3}+3}{2\sqrt{2}}\right)$

$\Rightarrow x^{\text{'}}=x\cos \alpha +y\sin \alpha ;y^{\text{'}}=-x\sin \alpha +y\cos \alpha$

$\frac{3\sqrt{3}-1}{2\sqrt{2}}=1\cos \alpha +2\sin \alpha ...\left(1\right)$

$\frac{\sqrt{3}+3}{2\sqrt{2}}=-1\sin \alpha +2\cos \alpha ...\left(2\right)$

Now, adding equations $\left(1\right)+\left(2\right)$,

$\frac{3\sqrt{3}-1}{2\sqrt{2}}+2\left(\frac{\sqrt{3}+3}{2\sqrt{2}}\right)=\cos \alpha +2\sin \alpha -2\sin \alpha +4\cos \alpha$

$\Rightarrow \frac{3\sqrt{3}-1+2\sqrt{3}+6}{2\sqrt{2}}=5\cos \alpha$

$\Rightarrow \frac{5\sqrt{3}+5}{2\sqrt{2}}=5\cos \alpha$

$\Rightarrow \frac{\sqrt{3}+1}{2\sqrt{2}}=\cos \alpha$

$\Rightarrow \cos 15°=\cos \alpha$

$\therefore \alpha =15°=\frac{\mathrm{\pi }}{12}$