1. Identify the interface and the normal

The boundary is the $XY$-plane, i.e. $z=0$.

So, the normal to the interface is along the $z$-axis. Since the ray is coming from $M_1$ ($z\ge 0$) toward $M_2$ ($z<0$), the relevant normal direction is along $-\hat{k}$.

2. Given refractive indices

$n_1=\sqrt{2}, \qquad n_2=\sqrt{3}$

3. Direction vector of the incident ray

$\vec P=4\sqrt{3}\,\hat i-3\sqrt{3}\,\hat j-5\,\hat k$

The angle of incidence is the angle between the ray direction and the normal.

4. Find the magnitude of $\vec P$

$|\vec P|=\sqrt{(4\sqrt{3})^2+(-3\sqrt{3})^2+(-5)^2}$

$=\sqrt{48+27+25}$

$=\sqrt{100}=10$

5. Find angle of incidence $i$

Using the normal along $-\hat k$,

$\cos i=\frac{\vec P\cdot(-\hat k)}{|\vec P|}$

Since $\vec P\cdot(-\hat k)=5$,

$\cos i=\frac{5}{10}=\frac12$

Hence,

$i=60^\circ$

Therefore,

$\sin i=\sin 60^\circ=\frac{\sqrt3}{2}$

6. Apply Snell's law

$n_1\sin i=n_2\sin r$

$\sqrt2\cdot \frac{\sqrt3}{2}=\sqrt3\sin r$

$\frac{\sqrt6}{2}=\sqrt3\sin r$

$\sin r=\frac{\sqrt6}{2\sqrt3}=\frac{\sqrt2}{2}$

So,

$r=45^\circ$

7. Difference between angle of incidence and refraction

$i-r=60^\circ-45^\circ=15^\circ$

Therefore, the required value is

$\boxed{15}$

8. Comparison with stored correct answer

Stored correct answer = $15$.

This matches the derived answer.