If you always want the $(x-x_1,y-y_1)$ vector to be rotated $90°$ counter-clockwise against the $(x_2-x_1,y_2-y_1)$ one, then you have
\begin{align*}
n(x-x_1) &= m(y_1-y_2) \\
n(y-y_1) &= m(x_2-x_1)
\end{align*}
So the left $y$ difference is the right $x$ difference, and the left $x$ difference is the negative of the right $y$ difference. This amounts to the $90°$ rotation I mentioned.
Solve the above and you find
\begin{align*}
x &= \frac{m(y_1-y_2)}n+x_1 \\
y &= \frac{m(x_2-x_1)}n+y_1
\end{align*}
which corresponds to one of your four solutions.
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 $ABC$.
Best Answer
Since the three lines intersect precisely at the orthocenter = the origin, we have that:
$$\text{intersection of lines I, III}\;:\;\begin{cases}2x+3y=1\\ ax+by=1\end{cases}\implies \begin{cases}\;2ax+3ay=a\\\!\!-2ax-2by=-2\end{cases}\implies$$
$$(3a-2b)y=a-2\implies y=\frac{a-2}{3a-2b}$$
And since we're given $\;y=0\implies a=2\;$ , and from here...continue.
You don't need all the equations for the intersections, but you need, as far as I can tell, at least one set of equations for one intersection point (which is known: the orthocenter)