Let $ABC$ be a triangle with $AB=3$ and $AC=4$. It is given that there does not exist a point $D$, different from $A$ and not lying on line $BC$, such that the Euler line of $ABC$ coincides with the Euler line of $DBC$. The square of the product of all possible lengths of $BC$ can be expressed in the form $m+n\sqrt p$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $100m+10n+p$.

Note: For this problem, consider every line passing through the center of an equilateral triangle to be an Euler line of the equilateral triangle. Hence, if $D$ is chosen such that $DBC$ is an equilateral triangle and the Euler line of $ABC$ passes through the center of $DBC$, then consider the Euler line of $ABC$ to coincide with "the" Euler line of $DBC$.

[i]Proposed by Michael Ren[/i]
1. **Understanding the Problem:**
   We are given a triangle \(ABC\) with \(AB = 3\) and \(AC = 4\). We need to find the square of the product of all possible lengths of \(BC\) such that there does not exist a point \(D\) (different from \(A\) and not on line \(BC\)) where the Euler line of \(\triangle ABC\) coincides with the Euler line of \(\triangle DBC\). The result should be expressed in the form \(m + n\sqrt{p}\), where \(m\), \(n\), and \(p\) are positive integers and \(p\) is not divisible by the square of any prime. Finally, we need to find \(100m + 10n + p\).

2. **Euler Line and Reflection:**
   The Euler line of a triangle passes through several important points, including the orthocenter \(H\) and the circumcenter \(O\). For the Euler lines of \(\triangle ABC\) and \(\triangle DBC\) to coincide, certain geometric conditions must be met. Specifically, the reflection of the Euler line of \(\triangle ABC\) across \(BC\) must intersect the circumcircle of \(\triangle ABC\) at two distinct points.

3. **Complex Number Representation:**
   To simplify calculations, we use complex numbers. Let \(A = a\), \(B = b\), and \(C = c\) be points on the unit circle. We set \(a = -bc\) to ensure that \(A\) lies on the unit circle. The reflection of the Euler line across \(BC\) must be tangent to the circumcircle at a point \(A'\).

4. **Solving for \(BC\):**
   Using the given lengths \(AB = 3\) and \(AC = 4\), we set up the equations:
   \[
   AB^2 = |a - b|^2 = 9 \quad \text{and} \quad AC^2 = |a - c|^2 = 16
   \]
   We also use the fact that the reflection of the Euler line must be tangent to the circumcircle at \(A'\), leading to the equation:
   \[
   b + c + \frac{1}{b} + \frac{1}{c} = 2
   \]
   This simplifies to:
   \[
   \cos(\beta) + \cos(\gamma) = 1
   \]
   where \(b = e^{i\beta}\) and \(c = e^{i\gamma}\).

5. **Calculating Cosines:**
   We solve the system of equations:
   \[
   4|a - b| = 3|a - c|
   \]
   Squaring both sides, we get:
   \[
   16(2 - a/b - b/a) = 9(2 - a/c - c/a)
   \]
   Substituting \(a = -bc\), we find:
   \[
   16(2 + c + 1/c) = 9(2 + b + 1/b)
   \]
   Simplifying, we get:
   \[
   16(1 + \cos(\gamma)) = 9(1 + \cos(\beta))
   \]
   Solving this system, we find:
   \[
   \cos(\beta) = \frac{23}{25} \quad \text{and} \quad \cos(\gamma) = \frac{2}{25}
   \]

6. **Finding \(BC\):**
   Using the cosine values, we calculate:
   \[
   BC^2 = |b - c|^2 = 2 - 2\cos(\beta - \gamma)
   \]
   Using the sine values:
   \[
   \sin(\beta) = \pm \frac{4\sqrt{6}}{25} \quad \text{and} \quad \sin(\gamma) = \pm \frac{3\sqrt{69}}{25}
   \]
   We find:
   \[
   BC^2 = 2 - 2\left(\frac{23}{25} \cdot \frac{2}{25} \pm \frac{4\sqrt{6}}{25} \cdot \frac{3\sqrt{69}}{25}\right)
   \]
   Simplifying, we get:
   \[
   BC^2 = \frac{1158}{625} \pm \frac{72\sqrt{46}}{625}
   \]

7. **Final Calculation:**
   The product of all possible lengths of \(BC\) is:
   \[
   \left(\frac{1158}{625} + \frac{72\sqrt{46}}{625}\right) \cdot \left(\frac{1158}{625} - \frac{72\sqrt{46}}{625}\right) = 49 + 588\sqrt{2}
   \]
   Therefore, \(m = 49\), \(n = 588\), and \(p = 2\).

The final answer is \( \boxed{ 100m + 10n + p = 100 \cdot 49 + 10 \cdot 588 + 2 = 4900 + 5880 + 2 = 10782 } \).