Given $\triangle ABC$ can we construct point $O$ such that $AO\times BC=BO\times AC=CO\times AB$? Given a $\triangle ABC$, is it possible to construct, with compass and straightedge, a point $O$ such that
$$AO\cdot BC=BO\cdot AC=CO\cdot AB$$
Does that point exist?
 A: These points are known
in ETC
as 1st and 2nd isodynamic points,
the triangle centers $X_{15}$ and $X_{16}$.
Isodynamic point:

In Euclidean geometry, the isodynamic points of a triangle are points
associated with the triangle, with the properties that  ... the
distances from the isodynamic point to the triangle vertices are
inversely proportional to the opposite side lengths of the triangle.

The barycentric coordinates of these points are
\begin{align} 
X_{15}:\quad&
a\sin(\alpha + \tfrac\pi3) &: b\sin(\beta + \tfrac\pi3) : c \sin(\gamma + \tfrac\pi3)
,\\
X_{16}:\quad&
a\sin(\alpha - \tfrac\pi3) &: b\sin(\beta - \tfrac\pi3) : c \sin(\gamma - \tfrac\pi3)
.
\end{align}
As a linear combination of the vertices,
\begin{align} 
X_{15}&=\frac{u\cdot A+v\cdot B+w\cdot C}{u+v+w}
,\\
u&=a\sin(\alpha + \tfrac\pi3)
,\\
v&=b\sin(\beta + \tfrac\pi3)
,\\
w&=c \sin(\gamma + \tfrac\pi3)
, 
\end{align}
and the invariants are
\begin{align} 
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2+4\sqrt3 S}}
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2-4\sqrt3 S}}
,
\end{align}
where $S$ is the area of $\triangle ABC$.
Example: for the the nominal $6-9-13$ triangle,
\begin{align} 
a&=6,\quad b=9,\quad c=13,\quad S=4\sqrt{35}
,\\
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
=\frac{702 \sqrt2}{\sqrt{286+16\sqrt{105}}}
\approx 46.80
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
=\frac{702 \sqrt2}{\sqrt{286-16\sqrt{105}}}
\approx 89.86
.
\end{align}
Construction.

Points $A_b.A_e$ and $B_b,B_e$
are the feet of the internal and external bisectors
of the angles $CAB=\alpha$
and $ABC=\beta$, respectively.
Points
$O_a=\tfrac12(A_b+A_e)$,
$O_b=\tfrac12(B_b+B_e)$
are the centers of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
through points
$A,A_b,A_e$
and $B,B_b,B_e$, respectively.
Intersection of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
gives the pair of isodynamic points,
1st, $X_{15}$ inside $\triangle ABC$
and 2nd, $X_{16}$, outside of $\triangle ABC$.
A: $\color{blue}{\text{Due to a nonstandard usage in the question,}}$
$\color{blue}{\text{it was assumed that the multiplication symbol referred to a vector cross product,}}$
$\color{blue}{\text{and this answer was constructed accordingly.}}$
$\color{blue}{\text{It would not apply if the multiplication is a product of lengths.}}$
Yes, provided you watch your signs.  You have to render the second cross product as $BO×\color{blue}{CA}$.
Let $O$ be the interior point.  Then $|AO×BC|$ measures twice the sum of $|\triangle AOB|$ and $|\triangle AOC|$ and the vector is directed into the region of space from which $A,B,C$ appear to be in clockwise order.  Cyclic permutations of this apply for $BO×CA$ and $CO×AB$.  Then the cross products $AO×BC,BO×CA,CO×AB$ are all identical by making the areas of the smaller triangles equal, which means $O$ is the centroid.
