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Solution

Given $P(5,4),\ Q(-2,4)$ and orthocentre $O\left(2,\frac{14}{5}\right)$

$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$

Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$

Using orthocentre condition (slopes), valid point is $R(2,-6)$

Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$

$c+2d=\frac{5}{3}+\frac{4}{3}=3$

$\boxed{3}$