In triangle $ABC$, $AB=3\sqrt{30}-\sqrt{10}$, $BC=12$, and $CA=3\sqrt{30}+\sqrt{10}$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $AC$. Denote $l$ as the line passing through the circumcenter $O$ and orthocenter $H$ of $ABC$, and let $E$ and $F$ be the feet of the perpendiculars from $B$ and $C$ to $l$, respectively. Let $l'$ be the reflection of $l$ in $BC$ such that $l'$ intersects lines $AE$ and $AF$ at $P$ and $Q$, respectively. Let lines $BP$ and $CQ$ intersect at $K$. $X$, $Y$, and $Z$ are the reflections of $K$ over the perpendicular bisectors of sides $BC$, $CA$, and $AB$, respectively, and $R$ and $S$ are the midpoints of $XY$ and $XZ$, respectively. If lines $MR$ and $NS$ intersect at $T$, then the length of $OT$ can be expressed in the form $\frac{p}{q}$ for relatively prime positive integers $p$ and $q$. Find $100p+q$.

[i]Proposed by Vincent Huang and James Lin[/i]
1. **Identify Key Points and Properties:**
   - Given triangle \(ABC\) with sides \(AB = 3\sqrt{30} - \sqrt{10}\), \(BC = 12\), and \(CA = 3\sqrt{30} + \sqrt{10}\).
   - \(M\) and \(N\) are midpoints of \(AB\) and \(AC\), respectively.
   - \(O\) is the circumcenter and \(H\) is the orthocenter of \(\triangle ABC\).
   - Line \(l\) passes through \(O\) and \(H\).
   - \(E\) and \(F\) are the feet of the perpendiculars from \(B\) and \(C\) to \(l\).
   - Line \(l'\) is the reflection of \(l\) in \(BC\) and intersects lines \(AE\) and \(AF\) at \(P\) and \(Q\), respectively.
   - Lines \(BP\) and \(CQ\) intersect at \(K\).
   - \(X\), \(Y\), and \(Z\) are the reflections of \(K\) over the perpendicular bisectors of sides \(BC\), \(CA\), and \(AB\), respectively.
   - \(R\) and \(S\) are the midpoints of \(XY\) and \(XZ\), respectively.
   - Lines \(MR\) and \(NS\) intersect at \(T\).

2. **Determine the Position of \(K\):**
   - \(K\) is the orthocenter of \(\triangle AHO\).
   - The isogonal conjugate of \(K\) lies on \(OH\), denote this point as \(J\).

3. **Use Isogonal Conjugates:**
   - Since \(AJ\) and \(AK\) are isogonal with respect to \(\angle HAO\), we can find \(OJ\).
   - The isogonal conjugate of \(K\) in \(\triangle ABC\) lies on \(OH\).

4. **Medial Triangle and Nine-Point Center:**
   - \(T\) in the medial triangle of \(\triangle ABC\) corresponds to \(J\) in \(\triangle ABC\).
   - The nine-point center \(N\) of \(\triangle ABC\) is the midpoint of \(OH\).

5. **Calculate \(OT\):**
   - The distance \(OJ\) can be found using the properties of isogonal conjugates.
   - Since \(T\) is in the medial triangle, the distance \(NT\) is half of \(OJ\).
   - Using the fact that \(\angle ABC\) is obtuse, we have \(ON - NT = OT\).

6. **Express \(OT\) in the Form \(\frac{p}{q}\):**
   - Calculate the exact value of \(OT\) and express it in the form \(\frac{p}{q}\) where \(p\) and \(q\) are relatively prime positive integers.

7. **Find \(100p + q\):**
   - Once \(OT\) is expressed as \(\frac{p}{q}\), compute \(100p + q\).

The final answer is \(\boxed{100p + q}\).