The area of a convex quadrilateral $ABCD$ is $18$. If $|AB|+|BD|+|DC|=12$, then what is $|AC|$?

$ \textbf{(A)}\ 9
\qquad\textbf{(B)}\ 6\sqrt 3
\qquad\textbf{(C)}\ 8
\qquad\textbf{(D)}\ 6
\qquad\textbf{(E)}\ 6\sqrt 2
$
1. Let \( AB = x \), \( BD = y \), and \( CD = z \). Given that the area of the quadrilateral \(ABCD\) is 18, we can use the formula for the area of a quadrilateral with perpendicular diagonals:
   \[
   \text{Area} = \frac{1}{2} \times d_1 \times d_2
   \]
   where \(d_1\) and \(d_2\) are the lengths of the diagonals. 

2. Given \( |AB| + |BD| + |DC| = 12 \), we have:
   \[
   x + y + z = 12
   \]

3. We know the area of \(ABCD\) is 18, so:
   \[
   18 = \frac{1}{2} \times AC \times BD
   \]
   Simplifying, we get:
   \[
   36 = AC \times BD
   \]

4. From the given condition \( x + y + z = 12 \), we can use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
   Substituting \( x + y + z = 12 \):
   \[
   4 \geq \sqrt[3]{xyz}
   \]
   Cubing both sides:
   \[
   64 \geq xyz
   \]

5. To find \(AC\), we need to consider the geometry of the quadrilateral. Since \(BD\) is perpendicular to both \(AB\) and \(AC\), we can infer that \(BD\) is the height from \(A\) to \(DC\) and from \(B\) to \(AC\).

6. Let the parallel line through \(A\) to \(BD\) meet \(DC\) at \(E\). Since \(AE = BD = 6\) and \(EC = AB + CD = 6\), we can use the Pythagorean theorem in triangle \(AEC\):
   \[
   AC = \sqrt{AE^2 + EC^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}
   \]

Conclusion:
\[
\boxed{6\sqrt{2}}
\]