Given: $\overrightarrow{A}=4\sqrt{3i}-3\sqrt{3j}-5\hat{k}$

As incident vector $A$ makes $i$ angle with normal $Z$-axis and refracted vector $R$ makes $r$ angle with normal $Z$- axis.

The angle made by the incident ray with the normal is given as 

$i={\cos }^{-1}\left(\frac{A_z}{A}\right)={\cos }^{-1}\left(\frac{5}{\sqrt{(4\sqrt{3}{)}^2+(3\sqrt{3}{)}^2+5^2}}\right)$

$={\cos }^{-1}\left(\frac{5}{10}\right)\Rightarrow i=60°$

Applying Snell's law of refraction at the interface of two media 

${\mu }_1\sin i={\mu }_2\sin r$

$\Rightarrow \sqrt{2}\sin 60°=\sqrt{3}\times \sin r$

$\Rightarrow r=45°$

Difference between $i$ and $r$ $=60°-45°=15°$