Barycentric coordinates provide a simple method. Let we assume that our triangle is $ABC$ and the side lengths are $a,b,c$. $a,b,c$ can be easily computed from the Pythagorean theorem, and the following barycentric coordinates
$$ I=[a,b,c],\qquad O=[a^2(b^2+c^2-a^2),b^2(a^2+c^2-b^2),c^2(a^2+b^2-c^2)]$$
$$ H=\left[\frac{1}{b^2+c^2-a^2},\frac{1}{a^2+c^2-b^2},\frac{1}{a^2+b^2-c^2}\right] $$
that we may summarize as $P=[p_a,p_b,p_c]$, give the vector identity
$$ P = \frac{p_a A+p_b B+p_c C}{p_a+p_b+p_c}$$
from which it is straightforward to compute the cartesian coordinates of $P$ from the coordinates of $A,B,C$. It is interesting to point out that Euler's theorem gives a further shortcut, since $3G=A+B+C=2O+H$ allows us to compute the coordinates of $H$ from the coordinates of $O$ (or the opposite) in a very simple way.
Here, as usual, $G,I,O,H$ stand for the centroid, incenter, circumcenter and orthocenter of a triangle. The derivation of their barycentric coordinates is straightforward from the computation of their trilinear coordinates, i.e. from the computation of their distances from the triangle sides in terms of $a,b,c$.
About your second question: up to reflections we may assume that $A$ lies on the positive $x$ axis, $B$ lies on the positive $y$ axis and $C$ lies on the positive $z$ axis. We are claiming that the projection of $O$ on the $ABC$-plane $\pi$ is given by the orthocenter of $ABC$. Let we denote such projection with $P$ and consider the plane $\pi_A$ through $O,A,P$. By minimality of $OP$, $\pi_A$ has to be orthogonal to the $BC$ line: otherwise, it would be possible to move a bit the $\pi_A$ plane and decrease the distance between $O$ and $\pi\cap\pi_A$. It follows that $\pi\cap\pi_A$ is orthogonal to $BC$, and by repeating the same argument we get that $P$ is the orthocenter of $ABC$.
If $G$ is the centroid of the triangle, that relation follows from
$$
\vec{AO} + \vec{BO} + \vec{CO}=
\vec{AG} + \vec{BG} + \vec{CG}+3\vec{GO}
=3\vec{GO}=\vec{HO}
$$
where the last one is a well-known equality, arising in the proof for the Euler line, see e.g. https://en.wikipedia.org/wiki/Euler_line#A_vector_proof .
Best Answer
As you are looking for a different way to solve your question, here is one.
Take a look at the following figure : it provides positions for $A,B,C$ that fullfill all the constraints :
Fig. 1.
How is it possible to obtain these points (out of which the strange result you are asked is easy to compute) ?
First of all, let us transform the implicit equation of straight line $BC$ into a parametric form ; as it passes through point $\binom{0}{1}$ with directing vector $\binom{3}{4}$, we can write :
$$\binom{x}{y}=\binom{0}{1}+t\binom{3}{4} \ \iff \ \begin{cases}x&=&3t\\y&=&1+4t\end{cases}\tag{1}$$
In particular, the coordinates of $B$ and $C$ are resp.
$$\begin{cases}x_B&=&3b\\y_B&=&1+4b\end{cases} \ \ \ \ \ \text{and} \ \ \ \ \begin{cases}x_C&=&3c\\y_C&=&1+4c\end{cases}\tag{2}$$
for specific values $b, \ c$ of parameter $t$.
Besides, $A$ belongs to the straight line passing by $H$ and orthogonal to $BC$ ; the parametric form of this straight line is easily found to be :
$$\binom{x}{y}=\binom{1}{2}+t\binom{4}{-3} \ \ \ \iff \ \ \ \begin{cases}x_A&=&1+4a\\y_A&=&2-3a\end{cases}\tag{3}$$
Now, let us locate the constraints : as we need three precise values for $a,b,c$, we need three constraints. Here they are :
$$PA^2=PB^2=PC^2 \ \ \ \ \text{and} \ \ \ \ \overrightarrow{AB} \perp \overrightarrow{CH},\tag{4}$$
giving rise, using (2) and (3), to the following system :
$$(1+4a-2)^2+(2-3a-3)^2=(3b-2)^2+(1+4b-3)^2=(3c-2)^2+(1+4c-3)^2$$ $$(3b-1)(1+4a-3c)+(1+4b-2)((2-3a)-(1+4c))=0\tag{5}$$
that is easily solved by a CAS giving the following values of parameters $a,b,c$ (please note that $b$ and $c$ can be exchanged) :
$$a=\frac{4}{25}, \ \ b=\frac{14+3\sqrt{6}}{25}, \ \ c=\frac{14-3\sqrt{6}}{25}\tag{6}$$
Here is the Matlab program that has given these values and Fig. 1 :
Remarks :
1) Relationships (4) take into account the characteristic properties
of the circumcenter (the unique point at equal distances from each vertex),
of the orthocenter (the intersection of 2 altitudes ; don't forget that we have previously considered $A$ to belong to the altitude opposite to $BC$).
2) There are other solutions for $a,b,c$ than (6) ; but none is satisfying ; some are with complex numbers (!), some others generating non acute triangles, and in particular a flat triangle.
3) $b$ and $c$ are roots of the same quadratic equation. It is very understandable because $B$ and $C$ play interchangeable roles.