12. (First "Xipeng Cup" - Grade 11) The vertices $Q, R$, and $S$ of the square $P Q R S$ slide along the sides $A B, B C, C A$ of an equilateral triangle $A B D$ with side length 2, respectively. Find the locus of point $P$.
Consider a general position (as shown in the figure).
Let $\angle C R S=\theta$,

then $\angle P R C=45^{\circ}$
$+\theta$, the distance from $P$ to $B C$
is $|P R| \sin \left(45^{\circ}+\theta\right)=\sqrt{2}|R S|$
$\cdot \sin \left(45^{\circ}+\theta\right)$. Let
$|R S|=x$, in $\triangle R S C, \triangle Q R S$,
we have $x(\sin \theta+\cos \theta)=3-\sqrt{3}$. (1)
On the other hand, the distance from $P$ to $B C$ (substituting (1) into
rii) $\sqrt{2} \times \sin \left(45^{\circ}+\theta\right)=3-\sqrt{3}$.
It can be seen that the distance from $P$ to $B C$ is a constant. After determining the boundary of the motion of point $P$, it is not difficult to find that the trajectory of point $P$ is a line segment parallel to $B C$, at a distance of $3-\sqrt{3}$ from $B C$, and on the same side of $B C$ as $A$. The length of this line segment is 2, and it is symmetric with respect to the median of side $B C$ in $\triangle A B C$.
[Elementary Number Theory]