BID for the angle

$\text{Using cosine rule}$

$\angle BAC=60^{\circ};\angle BCA =15^{\circ}$

$\angle DAB=30^{\circ}$ as AD is angle bisector.

$\angle ADB=\angle IDB~(of~course)=45^{\circ}$ , exterior angle property.

Now, we know in-center divides the angle bisector in the ratio $\large{\frac{\text{sum of sides excluding the side opposite to the vertex from which the angle bisector is drawn}}{\text{excluded side}}}$

$\large{\frac{AI}{ID}=\frac{3\sqrt{2}+\sqrt{6}+3\sqrt{2}-\sqrt{6}}{6}=\frac{\sqrt{2}}{1}=\frac{m}{n}}$

Using m-n cot theorem

$\large{(\sqrt{2}+1)k \cot(\angle BID)=k \cot (30^{\circ})-\sqrt{2}k \cot (45^{\circ})}$

$\large{\boxed{\angle BID=\cot^{-1} (\sqrt{6}+\sqrt{2}-\sqrt{3}-2)}}$

There are three steps to the solution of this problem on triangle.

1. Finding out $\Delta~XYZ \cong \Delta ABC$ so that converted data of $\Delta~XYZ$ is simpler than that of $\Delta~ABC$ .

2. The acute angle between angle bisector from X and Y = $\dfrac{\pi - Z} 2$ .

3. Use Cos Rule to find $\dfrac Z 2$ , and relate it to the answer reqired.

$\text{ 1.. Dividing all three sides of the } \triangle~ ABC ~by~\sqrt6,\\ we~get~a~similar~\Delta~XYZ.~\ \ \ \ \ ~M~ \cong~ D,~~ and~~ P~ \cong~ I .\\ So~\angle BID~=~\angle YPM.~~~~\color{#3D99F6}{XY=\sqrt3-1,~~YZ=\sqrt6,~~XZ=\sqrt3 +1.}\\~~\\$ $2.~~external~\angle YPM=internal~\angle s~\dfrac X 2+\dfrac Y 2.~~~\ \ \ \ ~\dfrac X 2+\dfrac Y 2+\dfrac Z 2=\dfrac \pi 2,\\ \implies \dfrac \pi 2 - \angle YPM=\dfrac Z 2.\\ \therefore~\color{#3D99F6}{CotBID}=CotYPM=Cot\{\dfrac \pi 2 - \dfrac Z 2\}=\color{#3D99F6}{Tan\dfrac Z 2}.\\~~\\$ $3.~~CosZ=\dfrac{(\sqrt6)^2+(\sqrt3 + 1)^2 - (\sqrt3 - 1)^2}{2*\sqrt6*(\sqrt3 +1)} =\dfrac{\sqrt3+2}{\sqrt2*(\sqrt3+1)}\\ But~CocZ=2*Cos^2\dfrac Z 2 -1,\ \ \ \ \ \ \implies~ Cos^2\dfrac Z 2=\dfrac{CosZ+1} 2\\ \therefore~Cos^2\dfrac Z 2=\dfrac{\sqrt3+2+\sqrt6+\sqrt2}{2*\sqrt2*(\sqrt3+1)}.\\ \therefore ~Tan^2\dfrac Z 2=\dfrac 1 { Cos^2\dfrac Z 2} - 1 =\dfrac {2*\sqrt2*(\sqrt3+1)} {\sqrt3+2+\sqrt6+\sqrt2}- 1.\\ On ~~rationalising~ CotBID= Tan\dfrac Z 2=\sqrt6+\sqrt2 - \sqrt3 - 2 \\ =\sqrt P+\sqrt Q - \sqrt R - S.\\\therefore~P+Q+R+S= \Huge\ \ \ \ \ \ \ \color{#D61F06}{13}$