The circumradius of acute triangle $ABC$ is twice of the distance of its circumcenter to $AB$. If $|AC|=2$ and $|BC|=3$, what is the altitude passing through $C$?

$ 
\textbf{(A)}\ \sqrt {14}
\qquad\textbf{(B)}\ \dfrac{3}{7}\sqrt{21}
\qquad\textbf{(C)}\ \dfrac{4}{7}\sqrt{21}
\qquad\textbf{(D)}\ \dfrac{1}{2}\sqrt{21}
\qquad\textbf{(E)}\ \dfrac{2}{3}\sqrt{14}
$
1. Let \( O \) be the circumcenter of triangle \( ABC \) and \( d \) be the distance from \( O \) to \( AB \). Let \( M \) be the midpoint of \( AB \). Given that the circumradius \( R \) is twice the distance from \( O \) to \( AB \), we have \( R = 2d \).

2. Since \( R = 2d \), triangle \( AOM \) is a \( 30^\circ-60^\circ-90^\circ \) triangle. This implies that \( \angle AOM = 30^\circ \) and \( \angle AOB = 120^\circ \) because \( \angle AOB = 2 \times \angle AOM \).

3. Using the Law of Cosines in triangle \( AOB \):
   \[
   AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle AOB)
   \]
   Given \( AC = 2 \), \( BC = 3 \), and \( \angle AOB = 120^\circ \), we have:
   \[
   AB^2 = 2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \cos(120^\circ)
   \]
   Since \( \cos(120^\circ) = -\frac{1}{2} \):
   \[
   AB^2 = 4 + 9 - 2 \cdot 2 \cdot 3 \cdot \left(-\frac{1}{2}\right)
   \]
   \[
   AB^2 = 4 + 9 + 6 = 19
   \]
   Therefore, \( AB = \sqrt{19} \).

4. To find \( d \), we use the relationship \( AB = 2d\sqrt{3} \):
   \[
   \sqrt{19} = 2d\sqrt{3}
   \]
   Solving for \( d \):
   \[
   d = \frac{\sqrt{19}}{2\sqrt{3}} = \frac{\sqrt{19}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{57}}{6}
   \]

5. The area of triangle \( ABC \) can be found using Heron's formula or directly using the sine rule. Here, we use the sine rule:
   \[
   [ABC] = \frac{1}{2} \cdot AC \cdot BC \cdot \sin(\angle ACB)
   \]
   Given \( \angle ACB = 60^\circ \):
   \[
   [ABC] = \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin(60^\circ) = \frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{\sqrt{3}}{2} = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}
   \]

6. The altitude from \( C \) to \( AB \) is given by:
   \[
   \text{Altitude} = \frac{2 \cdot [ABC]}{AB} = \frac{2 \cdot \frac{3\sqrt{3}}{2}}{\sqrt{19}} = \frac{3\sqrt{3}}{\sqrt{19}} = \frac{3\sqrt{57}}{19}
   \]

The final answer is \(\boxed{\frac{3\sqrt{57}}{19}}\)