Given a triangle $ABC$ with angles $\angle A = 60^{\circ}$, $\angle B = 75^{\circ}$, $\angle C = 45^{\circ}$, let $H$ be its orthocenter, and $O$ be its circumcenter. Let $F$ be the midpoint of side $AB$, and $Q$ be the foot of the perpendicular from $B$ onto $AC$. Denote by $X$ the intersection point of the lines $FH$ and $QO$. Suppose the ratio of the length of $FX$ and the circumradius of the triangle is given by $\frac{a + b \sqrt{c}}{d}$, then find the value of $1000a + 100b + 10c + d$.

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