Let $B = (20, 14)$ and $C = (18, 0)$ be two points in the plane. For every line $\ell$ passing through $B$, we color red the foot of the perpendicular from $C$ to $\ell$.  The set of red points enclose a bounded region of area $\mathcal{A}$. Find $\lfloor \mathcal{A} \rfloor$ (that is, find the greatest integer not exceeding $\mathcal A$).

[i]Proposed by Yang Liu[/i]
1. **Determine the distance \( BC \):**
   - The coordinates of points \( B \) and \( C \) are \( B = (20, 14) \) and \( C = (18, 0) \).
   - The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
     \[
     d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
     \]
   - Applying this formula to points \( B \) and \( C \):
     \[
     BC = \sqrt{(18 - 20)^2 + (0 - 14)^2} = \sqrt{(-2)^2 + (-14)^2} = \sqrt{4 + 196} = \sqrt{200} = 10\sqrt{2}
     \]

2. **Identify the set of red points:**
   - For every line \(\ell\) passing through \( B \), the foot of the perpendicular from \( C \) to \(\ell\) will trace out a circle centered at the midpoint of \( BC \) with radius equal to half the distance \( BC \).
   - The midpoint \( M \) of \( BC \) is:
     \[
     M = \left( \frac{20 + 18}{2}, \frac{14 + 0}{2} \right) = (19, 7)
     \]
   - The radius of the circle is:
     \[
     r = \frac{BC}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}
     \]

3. **Calculate the area of the circle:**
   - The area \( \mathcal{A} \) of a circle with radius \( r \) is given by:
     \[
     \mathcal{A} = \pi r^2
     \]
   - Substituting \( r = 5\sqrt{2} \):
     \[
     \mathcal{A} = \pi (5\sqrt{2})^2 = \pi \cdot 25 \cdot 2 = 50\pi
     \]

4. **Find the greatest integer not exceeding \( \mathcal{A} \):**
   - Using the approximation \( \pi \approx 3.14159 \):
     \[
     50\pi \approx 50 \times 3.14159 = 157.0795
     \]
   - The greatest integer not exceeding \( 157.0795 \) is \( 157 \).

The final answer is \( \boxed{157} \)