I am trying to solve a problem and got stuck in the following:
Point $P$ is a point inside triangle $ABC$ that satisfies the following conditions:
$$\angle BPC-\angle BAC=\angle APC-\angle ABC=\angle APB-\angle BCA$$ and $\angle BAC=60 \text{degrees}$ and length of $AP=12$ units.
What is the area of the triangle whose vertices are lowered from the point $P$ to the sides $AB, BC$, and $CA$?
I have no idea for this question.
Any triangle, the largest vertex angle of which is less than $120^o$, has a point $P$ in its interior that satisfies the conditions mentioned in OP’s problem statement. This point is known to triangle-geometers as $\text{X}_{15}\space 1^{\text{st}}\text{ Isodynamic Point}$ (see ETC). This is also one of the isogonal conjugates of the famous $\text{X}_{13}\space \text{Fermat Point}$. It lies on the Brocard axis of the reference triangle and its pedal triangle is equilateral (see $\mathrm{Fig.\space 2}$).
There are various constructions to locate this point in a given triangle, e.g. Wikipedia. $\mathrm{Fig.\space 1}$ shows a simple construction that is most appropriate to the scenario on hand. Triangle $ABC$ is the given scalene triangle and it is usually called the reference triangle by triangle-geometers. First calculate (or construct) the three angles, such that, $$\omega = 60^o+\measuredangle A\qquad\theta = 60^o+\measuredangle B\quad \text{and}\quad\phi = 60^o+\measuredangle C\qquad.$$
Now, construct three isosceles triangles $CVB$, $AUC$, and $BWA$, which have $\space\omega$, $\space\theta,\space$ and $\space\phi\space$ as their apex angles respectively. When we draw the circumcircles of these three isosceles triangles, they intersect at a common point, which is the sought point $P$.
Let us get back to OP’s question. Even with $\measuredangle A=60^o$, it is very difficult to determine the area of the mentioned triangle unless one assumes that $\triangle ABC$ is equilateral. That is exactly what @Piquito has done in his answer. Therefore, we are of the opinion that OP should accept that answer.
We are indebted to @YNK who has shown that this point is the remarkable (internal) isodynamic point $X_{15}$. See as well $X_{15}$ here in the ETC (Encyclopedia of Triangle Centers).
I will go further by using trilinear coordinates (and converting them to cartesian coordinates) in order to reach an explicit expression for the side $s$ of (equilateral !) triangle DEF made by the feet of the perpendiculars issued from point $X_{15}$ to the sides of $ABC$.
Knowing $s$, the area of triangle $DEF$ is easily obtained.
It is not the place to describe what are trilinear coordinates. It is enough to say that they are "cousin coordinates" of the better known barycentric coordinates with the following connection :
$$\text{(trilinear coord. )} \ (x,y,z) \ \leftrightarrow \ \text{(barycentric coord. )} (ax,by,cz) \tag{1}$$
where $a=BC,b=CA,d=AB$.
Therefore, trilinear coordinates and cartesian coordinates can be placed in correspondence as follows :
$$\text{(trilinear coord. )} \ (x,y,z) \ \ \leftrightarrow$$ $$\text{(cartesian coord. )} \ \frac{axA+bxB+cxC}{ax+by+cz}\tag{2}$$
Let $$\begin{cases} p&:=&\sin(\hat{A}+\pi/3)\\ q&:=&\sin(\hat{B}+\pi/3)\\ r&:=&\sin(\hat{C}+\pi/3) \end{cases}\tag{3}$$
The trilinear coordinates of $X_{15}, D, E, F$ are known to be (see the references mentionned at the beginning of this answer) resp. :
$$\begin{cases} X_{15} : &(p,&q,&r&)\\ D : &(0,&q+p\cos(\hat{C}),&r+p\cos(\hat{B})&)\\ E : &(p+q\cos(\hat{C}),&0,&r+q\cos(\hat{A})&)\\ F : &(p+r\cos(\hat{B}),&q+r\cos(\hat{A}),&0&) \end{cases}\tag{4}$$
With all these ingredients, we get :
$$D=\frac{b(q+p \cos(\hat{C}))B+c(r+p \cos(\hat{B}))C}{b(q+p \cos(\hat{C}))+c(r+p \cos(\hat{B}))}$$
$$E=\frac{a(p+q \cos(\hat{C}))A+b(r+q \cos(\hat{A}))C}{a(p+q \cos(\hat{C}))+b(r+q \cos(\hat{A}))}$$
which are resp. equivalent to :
$$\vec{CD}=\frac{b(q+p \cos(\hat{C}))}{b(q+p \cos(\hat{C}))+c(r+p \cos(\hat{B}))}\vec{CB}\tag{5}$$
$$\vec{CE}=\frac{a(p+q \cos(\hat{C}))}{a(p+q \cos(\hat{C}))+c(r+q \cos(\hat{A}))}\vec{CA}\tag{6}$$
By difference between (6) and (7), we are able to express $\vec{DE}$ (therefore its norm $s$, the looked-for sidelength) : as a function of known quantities :
$$\vec{DE}=\tfrac{a(p+q \cos(\hat{C}))}{a(p+q \cos(\hat{C}))+c(r+q \cos(\hat{A}))}\vec{CA}-\tfrac{b(q+p \cos(\hat{C}))}{b(q+p \cos(\hat{C}))+c(r+p \cos(\hat{B}))}\vec{CB}\tag{7}$$
In the Matlab program below which has generated Fig. 1, we have used complex quantities giving the value of $s$ as a side-result, instead of having it at once by using "big formula" (7).
Sanity check : if $ABC$ is equilateral, $p=q=r=\sqrt{3}/2$ ; a routine verification shows that (7) is reduced to $\vec{ DE}=\tfrac12 \vec{BA}$$.
clear all;close all;hold on;axis equal;axis([-1,1,-1,1]);axis off
% Vertices $A,B,C$ can be assumed WLOG to lay on the unit circle :
alpha=17*pi/12;beta=17*pi/9;gamma=11*pi/15; % for example
A=exp(i*alpha);
B=exp(i*beta);
C=exp(i*gamma);
V1=[A,B,C];
plot([V1,V1(1)],'r','linewidth',5);
a=abs(B-C);b=abs(C-A);c=abs(B-A);
Aa=pi+(gamma-beta)/2;
Ba=(alpha-gamma)/2;
Ca=(beta-alpha)/2;
p=sin(Aa+pi/3);
q=sin(Ba+pi/3);
r=sin(Ca+pi/3);
% transformation trilinear -> cartesian :
tri2car=@(t)((a*t(1)*A+b*t(2)*B+c*t(3)*C)/(a*t(1)+b*t(2)+c*t(3)));
Xc=tri2car([p,q,r]);
Dc=tri2car([0,q+p*cos(Ca),r+p*cos(Ba)]);
Ec=tri2car([p+q*cos(Ca),0,r+q*cos(Aa)]);
Fc=tri2car([p+r*cos(Ba),q+r*cos(Aa),0])
s=abs(Dc-Ec), % s = 0.798325880115807 in our case.
plot([Xc,Dc,Xc,Ec,Xc,Fc],'-ok');
V2=[Dc,Ec,Fc];
text(real(V1)+[-0.1,0.1,-0.1],imag(V1)+[-0.1,0,0.1],{'A','B','C'})
text(real(V2)+[0.05,-0.15,0],imag(V2)+[0.1,0,-0.1],{'D','E','F'})
text(real(Xc)+0.05,imag(Xc),'X')
plot([V2,V2(1)],'g','linewidth',5)
HINT.-$$\angle{BPC}+\angle{APC}+\angle{APB}=360^{\circ}\\\angle{BAC}+\angle{ABC}+\angle{ACB}=180^{\circ}$$ This suggest the answer $\triangle{ABC}$ equilateral so, by property of medians, of height equal to $12+\dfrac{12}{2}=18$ and side $12\sqrt3$.
Then the triangle of required area is the equilateral triangle of side $6\sqrt3$.
