If $ \left|BC\right| \equal{} a$, $ \left|AC\right| \equal{} b$, $ \left|AB\right| \equal{} c$, $ 3\angle A \plus{} \angle B \equal{} 180{}^\circ$ and $ 3a \equal{} 2c$, then find $ b$ in terms of $ a$.

$\textbf{(A)}\  \frac {3a}{2} \qquad\textbf{(B)}\  \frac {5a}{4} \qquad\textbf{(C)}\  a\sqrt {2} \qquad\textbf{(D)}\  a\sqrt {3} \qquad\textbf{(E)}\ \frac {2a\sqrt {3} }{3}$
1. Given the conditions:
   - \( |BC| = a \)
   - \( |AC| = b \)
   - \( |AB| = c \)
   - \( 3\angle A + \angle B = 180^\circ \)
   - \( 3a = 2c \)

2. Let \( \angle A = \alpha \) and \( \angle B = \beta \). From the given condition \( 3\alpha + \beta = 180^\circ \), we can express \(\beta\) in terms of \(\alpha\):
   \[
   \beta = 180^\circ - 3\alpha
   \]

3. Since the sum of the angles in a triangle is \(180^\circ\), we have:
   \[
   \alpha + \beta + \angle C = 180^\circ
   \]
   Substituting \(\beta\):
   \[
   \alpha + (180^\circ - 3\alpha) + \angle C = 180^\circ
   \]
   Simplifying:
   \[
   \alpha + 180^\circ - 3\alpha + \angle C = 180^\circ
   \]
   \[
   -2\alpha + \angle C = 0
   \]
   \[
   \angle C = 2\alpha
   \]

4. Now, consider the triangle \( \triangle ABC \). We are given that \( 3a = 2c \), so:
   \[
   c = \frac{3a}{2}
   \]

5. Let \( Q \) be a point on \( AB \) such that \( CQ \) bisects \( \angle ACB \). Since \( \angle ACB = 2\alpha \), \( \angle ACQ = \alpha \) and \( \angle BCQ = \alpha \).

6. By the Angle Bisector Theorem, we have:
   \[
   \frac{AQ}{QB} = \frac{AC}{BC} = \frac{b}{a}
   \]

7. Let \( AQ = x \) and \( QB = y \). Then:
   \[
   \frac{x}{y} = \frac{b}{a}
   \]
   Since \( AQ + QB = AB = c = \frac{3a}{2} \), we have:
   \[
   x + y = \frac{3a}{2}
   \]
   Substituting \( y = \frac{ax}{b} \):
   \[
   x + \frac{ax}{b} = \frac{3a}{2}
   \]
   \[
   x \left(1 + \frac{a}{b}\right) = \frac{3a}{2}
   \]
   \[
   x \left(\frac{b + a}{b}\right) = \frac{3a}{2}
   \]
   \[
   x = \frac{3a}{2} \cdot \frac{b}{b + a}
   \]

8. Since \( AQ = x \) and \( QC = \frac{2b}{3} \) (from the similarity of triangles), we have:
   \[
   \frac{2b}{3} = \frac{3a}{2} - \frac{2a}{3}
   \]
   Simplifying:
   \[
   \frac{2b}{3} = \frac{9a - 4a}{6}
   \]
   \[
   \frac{2b}{3} = \frac{5a}{6}
   \]
   \[
   2b = \frac{5a}{2}
   \]
   \[
   b = \frac{5a}{4}
   \]

The final answer is \(\boxed{\frac{5a}{4}}\)