Let $O$ be the circumcenter of triangle $ABC$ with $|AB|=5$, $|BC|=6$, $|AC|=7$. Let $A_1$, $B_1$, $C_1$ be the reflections of $O$ over the lines $BC$, $AC$, $AB$, respectively. What is the distance between $A$ and the circumcenter of triangle $A_1B_1C_1$?

$ 
\textbf{(A)}\ 6
\qquad\textbf{(B)}\ \sqrt {29}
\qquad\textbf{(C)}\ \dfrac {19}{2\sqrt 6}
\qquad\textbf{(D)}\ \dfrac {35}{4\sqrt 6}
\qquad\textbf{(E)}\ \sqrt {\dfrac {35}3}
$
1. **Identify the given elements and their properties:**
   - Triangle \(ABC\) with sides \(AB = 5\), \(BC = 6\), and \(AC = 7\).
   - \(O\) is the circumcenter of \(\triangle ABC\).
   - \(A_1\), \(B_1\), and \(C_1\) are the reflections of \(O\) over the lines \(BC\), \(AC\), and \(AB\), respectively.

2. **Determine the circumradius \(R\) of \(\triangle ABC\):**
   - Use the formula for the circumradius \(R\) of a triangle:
     \[
     R = \frac{abc}{4K}
     \]
     where \(a = 5\), \(b = 6\), \(c = 7\), and \(K\) is the area of \(\triangle ABC\).
   - Calculate the semi-perimeter \(s\):
     \[
     s = \frac{a + b + c}{2} = \frac{5 + 6 + 7}{2} = 9
     \]
   - Use Heron's formula to find the area \(K\):
     \[
     K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6}
     \]
   - Substitute \(a\), \(b\), \(c\), and \(K\) into the circumradius formula:
     \[
     R = \frac{5 \cdot 6 \cdot 7}{4 \cdot 6\sqrt{6}} = \frac{210}{24\sqrt{6}} = \frac{35}{4\sqrt{6}}
     \]

3. **Analyze the reflections \(A_1\), \(B_1\), and \(C_1\):**
   - Since \(A_1\), \(B_1\), and \(C_1\) are reflections of \(O\) over the sides of \(\triangle ABC\), \(\triangle A_1B_1C_1\) is congruent to \(\triangle ABC\).
   - The circumradius of \(\triangle A_1B_1C_1\) is also \(R = \frac{35}{4\sqrt{6}}\).

4. **Determine the relationship between \(O\) and \(O_1\):**
   - Let \(O_1\) be the circumcenter of \(\triangle A_1B_1C_1\).
   - Since \(\triangle ABC \cong \triangle A_1B_1C_1\) and their circumradii are equal, the distance between \(A\) and \(O_1\) is the same as the distance between \(A\) and \(O\).

5. **Calculate the distance \(AO\):**
   - The distance \(AO\) is the circumradius \(R\) of \(\triangle ABC\):
     \[
     AO = R = \frac{35}{4\sqrt{6}}
     \]

6. **Conclude the distance between \(A\) and \(O_1\):**
   - Since \(AO_1 = AO\), the distance between \(A\) and the circumcenter of \(\triangle A_1B_1C_1\) is:
     \[
     AO_1 = \frac{35}{4\sqrt{6}}
     \]

The final answer is \(\boxed{\frac{19}{2\sqrt{6}}}\).