Let $ABC$ be a triangle with $AB=2$ and $AC=3$. Let $H$ be the orthocenter, and let $M$ be the midpoint of $BC$. Let the line through $H$ perpendicular to line $AM$ intersect line $AB$ at $X$ and line $AC$ at $Y$. Suppose that lines $BY$ and $CX$ are parallel. Then $[ABC]^2=\frac{a+b\sqrt{c}}{d}$ for positive integers $a,b,c$ and $d$, where $\gcd(a,b,d)=1$ and $c$ is not divisible by the square of any prime. Compute $1000a+100b+10c+d$.

[i]Proposed by Luke Robitaille
1. **Identify the orthocenter and midpoints:**
   - Given triangle \(ABC\) with \(AB = 2\) and \(AC = 3\).
   - Let \(H\) be the orthocenter of \(\triangle ABC\).
   - Let \(M\) be the midpoint of \(BC\).

2. **Determine the line through \(H\) perpendicular to \(AM\):**
   - The line through \(H\) perpendicular to \(AM\) intersects \(AB\) at \(X\) and \(AC\) at \(Y\).

3. **Given condition:**
   - Lines \(BY\) and \(CX\) are parallel.

4. **Reframe the problem using the orthocenter properties:**
   - Note that \(A\) is the orthocenter of \(\triangle HBC\).
   - Let \(D, E, F\) be the feet of the altitudes from \(H, B, C\) respectively.
   - \(DH, BE, CF\) concur at \(A\).

5. **Use Brocard's Theorem:**
   - Consider cyclic quadrilateral \(BCEF\) with center \(M\).
   - By Brocard's Theorem, \(MA\) is perpendicular to \(HG\).

6. **Projective geometry application:**
   - Since \(DH, BE, CF\) concur, \((D, BC \cap EF; B, C) = (DG; BC) = -1\).
   - Projecting through \(A\), \((GH; XY) = -1\).

7. **Angle bisector property:**
   - Since \(\angle GDH = 90^\circ\), either \(DG\) or \(DH\) is the angle bisector of \(\angle XDY\).
   - Thus, \(\angle BDY = \angle CDX\).

8. **Midpoint and perpendicularity:**
   - Projecting \((DG; BC)\) through \(Y\), \((X, DY \cap XC; P_\infty, C) = -1\).
   - \(C\) is the midpoint of \(X\) and \(DY \cap XC = K\).
   - \(\angle CDX = \angle CDK\), so \(DXK\) is isosceles with altitude \(DC\).
   - \(DC \perp CX\) and \(DB \perp BY\).

9. **Homothety and bisector:**
   - \(DH \parallel XK\), so a homothety centered at \(Y\) implies \(A\) bisects \(DH\).

10. **Rephrased problem:**
    - Acute triangle \(HBC\) with orthocenter \(A\) satisfies \(AB = 2\) and \(AC = 3\).
    - \(A\) bisects the altitude from \(H\) to \(BC\).

11. **Computation:**
    - Using the properties and given lengths, compute the area of \(\triangle ABC\).
    - The area squared is given by \(\frac{a + b\sqrt{c}}{d}\).

12. **Final calculation:**
    - The area squared is \(\frac{13\sqrt{601} + 263}{72}\).

The final answer is \( \boxed{ 1000a + 100b + 10c + d = 270382 } \).