Let $ABC$ be a triangle such that $|BC|=7$ and $|AB|=9$. If $m(\widehat{ABC}) = 2m(\widehat{BCA})$, then what is the area of the triangle?

$ 
\textbf{(A)}\ 14\sqrt 5
\qquad\textbf{(B)}\ 30
\qquad\textbf{(C)}\ 10\sqrt 6
\qquad\textbf{(D)}\ 20 \sqrt 2
\qquad\textbf{(E)}\ 12 \sqrt 3
$
1. **Define the angles and sides:**
   Let $\angle BCA = \theta$ and $\angle ABC = 2\theta$. Since the sum of angles in a triangle is $180^\circ$, we have:
   \[
   \angle BAC = 180^\circ - \angle ABC - \angle BCA = 180^\circ - 2\theta - \theta = 180^\circ - 3\theta
   \]

2. **Use the Law of Sines:**
   According to the Law of Sines, we have:
   \[
   \frac{BC}{\sin(\angle BAC)} = \frac{AB}{\sin(\angle BCA)} = \frac{AC}{\sin(\angle ABC)}
   \]
   Substituting the known values:
   \[
   \frac{7}{\sin(180^\circ - 3\theta)} = \frac{9}{\sin(\theta)} = \frac{AC}{\sin(2\theta)}
   \]
   Since $\sin(180^\circ - x) = \sin(x)$, we get:
   \[
   \frac{7}{\sin(3\theta)} = \frac{9}{\sin(\theta)} = \frac{AC}{\sin(2\theta)}
   \]

3. **Express $\sin(3\theta)$ and $\sin(2\theta)$:**
   Using the triple angle and double angle formulas:
   \[
   \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)
   \]
   \[
   \sin(2\theta) = 2\sin(\theta)\cos(\theta)
   \]

4. **Relate the sides using the Law of Cosines:**
   Using the Law of Cosines on $\triangle ABC$:
   \[
   AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)
   \]
   Substituting the known values:
   \[
   AC^2 = 9^2 + 7^2 - 2 \cdot 9 \cdot 7 \cdot \cos(2\theta)
   \]
   \[
   AC^2 = 81 + 49 - 126 \cos(2\theta)
   \]
   \[
   AC^2 = 130 - 126 \cos(2\theta)
   \]

5. **Find $\cos(2\theta)$:**
   Using the double angle formula for cosine:
   \[
   \cos(2\theta) = 2\cos^2(\theta) - 1
   \]
   Let $\cos(\theta) = x$, then:
   \[
   \cos(2\theta) = 2x^2 - 1
   \]
   Substituting into the equation for $AC^2$:
   \[
   AC^2 = 130 - 126(2x^2 - 1)
   \]
   \[
   AC^2 = 130 - 252x^2 + 126
   \]
   \[
   AC^2 = 256 - 252x^2
   \]

6. **Solve for $AC$:**
   From the Law of Sines, we have:
   \[
   \frac{7}{\sin(3\theta)} = \frac{9}{\sin(\theta)}
   \]
   \[
   7 \sin(\theta) = 9 \sin(3\theta)
   \]
   \[
   7 \sin(\theta) = 9 (3 \sin(\theta) - 4 \sin^3(\theta))
   \]
   \[
   7 \sin(\theta) = 27 \sin(\theta) - 36 \sin^3(\theta)
   \]
   \[
   36 \sin^3(\theta) = 20 \sin(\theta)
   \]
   \[
   36 \sin^2(\theta) = 20
   \]
   \[
   \sin^2(\theta) = \frac{5}{9}
   \]
   \[
   \sin(\theta) = \frac{\sqrt{5}}{3}
   \]

7. **Find $\cos(\theta)$:**
   Using $\sin^2(\theta) + \cos^2(\theta) = 1$:
   \[
   \left(\frac{\sqrt{5}}{3}\right)^2 + \cos^2(\theta) = 1
   \]
   \[
   \frac{5}{9} + \cos^2(\theta) = 1
   \]
   \[
   \cos^2(\theta) = \frac{4}{9}
   \]
   \[
   \cos(\theta) = \frac{2}{3}
   \]

8. **Calculate the area:**
   Using the area formula for a triangle:
   \[
   \text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC)
   \]
   \[
   \text{Area} = \frac{1}{2} \cdot 9 \cdot 7 \cdot \sin(2\theta)
   \]
   \[
   \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{4\sqrt{5}}{9}
   \]
   \[
   \text{Area} = \frac{1}{2} \cdot 9 \cdot 7 \cdot \frac{4\sqrt{5}}{9} = \frac{1}{2} \cdot 7 \cdot 4\sqrt{5} = 14\sqrt{5}
   \]

The final answer is $\boxed{14\sqrt{5}}$