Let $ABC$ be a triangle with $AB = 6$, $AC = 8$, $BC = 7$. Let $H$ be the orthocenter of $\triangle ABC$. Let $D \neq H$ be a point on $\overline{AH}$ such that $\angle HBD = \frac{3}{2} \angle CAB + \frac{1}{2} \angle ABC - \frac{1}{2} \angle BCA$. Find $DH$.

Let's think step by step and output the final answer within \boxed{}.
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