To solve the problem as stated, it's not necessary to write equations for the altitudes, or even to find point $C$. It just asks for equations representing the three sides of the triangle.
Side $AB$ is the simplest case, since points $A$ and $B$ are given. You have already computed its slope $m_{AB}=4$, and can use either $A$ or $B$ as the point on the line.
For side $AC$, notice that it must be perpendicular to the line $BN$. We don't need to know where these lines intersect to conclude that $m_{AC}m_{BN} = -1$. And $m_{BN}$ is simple to compute from the two given points $B$ and $N$. Knowing the slope and a point $A$ on the line, you can write the equation for $AC$.
Similarly, side $BC$ is perpendicular to the line $AN$, so you can compute $m_{AN}$ and from that determine $m_{BC}$, then write the equation for $BC$.
(See figure below)
Here is a solution bypassing the obtention of vertices' coordinates, by using "pencils of lines".
Being given two lines $L_1$ and $L_2$ with resp. equations $$\begin{cases}u_1x+v_1y+w_1=0\\u_2x+v_2y+w_2=0\end{cases}$$
Any line passing through point $L_1 \cap L_2$ has general equation
$$m(u_1x+v_1y+w_1=0)+(1-m)(u_2x+v_2y+w_2=0)=0, \ \ m \in \mathbb{R}.$$
The set of all these lines is called the "pencil of lines" defined by $L_1$ and $L_2$.
Thus, the pencil of lines defined by $x+y-1=0$ and $2x+3y-6=0$ is:
$$a(x+y-1)+(1-a)(2x+3y-6)=0$$
$$\tag{1} \iff \ (2-a)x+(3-2a)y+(5a-6)=0$$
Among these lines, one is the height. This height is characterized by the fact that its normal vector is orthogonal to the normal vector of the third line:
$$\binom{2-a}{3-2a} \perp \binom{4}{-1} \ \ \iff \ \ (2-a)4+(3-2a)(-1)=0 \ \ \iff \ \ a=\frac52$$
(recall: a line with equation $ux+vy+w=0$ has normal vector $\binom{u}{v}.$)
Plugging this value of $a$ in (1) gives the equation of the height:
$$\tag{2}x+4y-13=0.$$
Working in the same way with a group of 2 other sides:
$$b(2x+3y-6)+(1-b)(4x-y+4)=0$$
$$\tag{3} \iff \ \ (4-2b)x+(4b-1)y+(4-10b)=0$$
$$\binom{4-2b}{4b-1} \perp \binom{1}{1} \ \ \iff \ \ (4-2b)1+(4b-1)1=0 \ \ \iff \ \ b=-\frac32$$
Plugging this value of $b$ in (3) gives the equation of the second height:
$$\tag{4}7x-7y+19=0.$$
The solution of the system formed by (2) and (4) are the coordinates of the orthocentre
$$H\left(\frac{3}{7},\frac{22}{7}\right).$$
Best Answer
The last point shares its $x$-coordinate with the first point and its $y$-coordinate with the second. Therefore the triangle is right, and right triangles have their orthocentres at the right angle – $\left(2,-\frac12\right)$ here.