Revisiting coordinate 3!

The equation of $AB$ is $ax+by+c=0$ where $a+c=2b$ . Since $A$ lies on this line, $a+b+c=0$ . Hence $b=0$ , and so $AB$ has equation $x=1$ . Thus $B$ has coordinates $(1,u)$ for some $u$ . Since $AB$ is parallel to the $y$ -axis, $HC$ is parallel to the $x$ -axis, and hence $C$ has coordinates $(v,4)$ for some $v$ .

Since $AH$ has gradient $3$ , $BC$ has gradient $-\tfrac13$ , and hence has equation $x + 3y = 3u+1$ . Thus $1 + -(3u+1) = 2\times3$ , and hence $u=-2$ . In addition $v + 12 = 3u+1 = -5$ , and hence $v = -17$ . Thus $A,B,C$ have coordinates $(1,1)$ , $(1,-2)$ , $(-17,4)$ respectively. Taking the average of these, the centroid $G$ of the triangle has coordinates $(-5,1)$ . Since $G$ lies on the line $OH$ with $OG = \tfrac13OH$ , we deduce that $O$ has coordinates $\big(-\tfrac{17}{2},-\tfrac12\big)$ . Thus $d+e = -\tfrac{17}{2} - \tfrac12 = \boxed{-9}$ .