There are three triangles. Two compose the largest one. I'll call the leftmost part $L,$ its counterpart on the right $R,$ and their composite $K.$
Looking at $R,$ we can write down an equation for our goal $x_2$ as follows: $$x_2^2=a_1^2+a_2^2-2a_1a_2\cos E.$$ Thus we only need find $a_1$ and $a_2,$ and we'd be done.
To find $a_1,$ look at $L,$ the we have that $$\frac{\sin B}{\text{Length}}=\frac{\sin A}{a_1}.$$ Similarly, by looking at $K,$ we find that $$\frac{\sin D}{\text{Length}}=\frac{\sin A}{a_2}.$$ This helps us to find $a_2.$
From the last two equations we eliminate $A$ to find an equation in $a_1,\,a_2.$
Now look at the triangle $R,$ from which we may write down $$\frac{\sin E}{x_2}=\frac{\sin D}{a_1}.$$
Together with the first equation, we find a $3×3$ system in $a_1,\,a_2\,x_2,$ which we can now solve for $x_2.$
OK, let's do this together. From the two middle equations we find that $$\sin A=\frac{a_1\sin B}{\text{Length}}=\frac{a_2\sin D}{\text{Length}},$$ so that $$a_1=\frac{\sin D}{\sin B}a_2.$$ From the last equation we also have $$a_1=\frac{\sin D}{\sin E}x_2,$$ so that we have that $$a_2=\frac{\sin D}{\sin E}\frac{\sin B}{\sin D}x_2=\frac{\sin B}{\sin D}x_2.$$
Substituting now into the first equation and solving for $x_2$ is straightforward, and gives $$x_2^2\left(\frac{\sin B+\sin D}{\sin E}+1\right)=\frac{2\sin B\sin D\cot E}{\sin E},$$ amd finally recall that $x_2>0.$
You already got $\sin\angle{BAC}=\frac{\sqrt 3+1}{2\sqrt 2}$ which can be written as
$$\sin{\angle{BAC}}=\frac{1}{\sqrt 2}\cdot\frac{\sqrt 3}{2}+\frac{1}{\sqrt 2}\cdot \frac 12=\sin(45^\circ+30^\circ)$$
which implies that
$$\angle{BAC}=75^\circ\quad\text{or}\quad 105^\circ$$
Now note that $$BC^2-AB^2-CA^2=2(2\sqrt 3-3)\gt 0$$which implies that $\angle{BAC}$ is obtuse, so $\angle{BAC}=\color{red}{105^\circ}$.
Best Answer
$\Delta BDC \sim \Delta BCA$ because $m( \angle BCD) = m( \angle BAC)$ and $\angle B$ is a common angle.
And we can continue from here, to end up with $AB = 16, AC=8$, which means that our answer is $\frac {7+8+6}{6+9+12}=\frac {7}{9}$
You can say the same thing using trignometric functions as: $$\frac {6}{\sin(\angle B)}= \frac {9}{\sin(\angle BCD)} \\ \frac {AC}{\sin(\angle B)}= \frac{12}{\sin(\angle A)} \\ m(\angle A) = m (\angle BCD), m(\angle C) = m(\angle BDC) \\ \implies \frac {6}{AC}=\frac{9}{12}=\frac{12}{AB}$$ And from that we can calculate the missing values: $AC = 8, AB = 16$. And then we can find the answer: $\frac {7+8+6}{6+9+12}=\frac {7}{9}$.