Find $x^2$

Let the side length of $\triangle ABC$ be $a$ and the coordinates be $B(0,0)$ , $A \left(\frac a2, \frac {\sqrt 3a}2\right)$ , and $D(u,v)$ .

To find $a$ and hence the coordinates of $A(u,v)$ , let $\angle DBC = \theta$ and applying cosine rule on $\triangle BCD$ :

$a^2 = 6^2 + 3^2 - 2(6)(3) \left(-\dfrac {\sqrt 3}2\right) = 45 + 18\sqrt 3$

To find the coordinates of $D(u,v)$ , let $\angle DBC = \theta$ and applying sine rule on $\triangle BCD$ :

$\begin{aligned} \frac {\sin \theta}3 & = \frac {\sin (150^\circ)}a = \frac 1{2a} \\ \implies \sin \theta & = \frac 3{2a} \\ \implies u & = 6\cos \theta = 6\sqrt{1-\frac 9{4a^2}} = \frac 3a \sqrt{4a^2 - 9} \\ & = \frac 3a \sqrt{4(45 + 18\sqrt 3)-9} = \frac 9a \sqrt{19+8\sqrt 3} \\ & = \frac 9a \sqrt{(4+\sqrt 3)^2} \frac {9(4+\sqrt 3)}a \\ \implies au & = 36 + 9\sqrt 3 \\ v & = 6\sin \theta = 6\left(\frac 3{2a}\right) = \frac 9a \\ \implies av & = 9 \end{aligned}$

By Pythagorean theorem :

$\begin{aligned} x^2 & = \left(u-\frac a2\right)^2 + \left(v-\frac {\sqrt 3a}2\right)^2 \\ & = u^2 - au+\frac {a^2}4 + v^2 - \sqrt 3av + \frac {3a^2}4 \\ & = u^2 + v^2 - au - \sqrt 3 av + a^2 \\ & = 6^2(\cos^2 \theta + \sin^2 \theta) - 36-9\sqrt 3 - 9\sqrt 3 + 45+18\sqrt 3 \\ & = 36 - 36 - 9\sqrt 3 - 9\sqrt 3 + 45 + 18\sqrt 3 \\ & = \boxed {45} \end{aligned}$