Let $ A \equal{} (0,9)$ and $ B \equal{} (0,12)$. Points $ A'$ and $ B'$ are on the line $ y \equal{} x$, and $ \overline{AA'}$ and $ \overline{BB'}$ intersect at $ C \equal{} (2,8)$. What is the length of $ \overline{A'B'}$?

$ \textbf{(A)}\  2 \qquad \textbf{(B)}\  2\sqrt2 \qquad \textbf{(C)}\  3 \qquad \textbf{(D)}\  2 \plus{} \sqrt 2\qquad \textbf{(E)}\ 3\sqrt 2$
1. **Identify the coordinates of points \( A \) and \( B \):**
   \[
   A = (0, 9) \quad \text{and} \quad B = (0, 12)
   \]

2. **Determine the coordinates of points \( A' \) and \( B' \) on the line \( y = x \):**
   Let \( A' = (a, a) \) and \( B' = (b, b) \).

3. **Find the equations of lines \( \overline{AA'} \) and \( \overline{BB'} \):**
   - The slope of \( \overline{AA'} \) is:
     \[
     \text{slope of } \overline{AA'} = \frac{a - 9}{a - 0} = \frac{a - 9}{a}
     \]
     The equation of line \( \overline{AA'} \) is:
     \[
     y - 9 = \frac{a - 9}{a} (x - 0) \implies y = \frac{a - 9}{a} x + 9
     \]
   - The slope of \( \overline{BB'} \) is:
     \[
     \text{slope of } \overline{BB'} = \frac{b - 12}{b - 0} = \frac{b - 12}{b}
     \]
     The equation of line \( \overline{BB'} \) is:
     \[
     y - 12 = \frac{b - 12}{b} (x - 0) \implies y = \frac{b - 12}{b} x + 12
     \]

4. **Find the intersection point \( C = (2, 8) \) of \( \overline{AA'} \) and \( \overline{BB'} \):**
   - Substitute \( x = 2 \) and \( y = 8 \) into the equation of \( \overline{AA'} \):
     \[
     8 = \frac{a - 9}{a} \cdot 2 + 9 \implies 8 = \frac{2a - 18}{a} + 9 \implies 8a = 2a - 18 + 9a \implies 8a = 11a - 18 \implies 3a = 18 \implies a = 6
     \]
   - Substitute \( x = 2 \) and \( y = 8 \) into the equation of \( \overline{BB'} \):
     \[
     8 = \frac{b - 12}{b} \cdot 2 + 12 \implies 8 = \frac{2b - 24}{b} + 12 \implies 8b = 2b - 24 + 12b \implies 8b = 14b - 24 \implies 6b = 24 \implies b = 4
     \]

5. **Determine the coordinates of \( A' \) and \( B' \):**
   \[
   A' = (6, 6) \quad \text{and} \quad B' = (4, 4)
   \]

6. **Calculate the length of \( \overline{A'B'} \):**
   Using the distance formula:
   \[
   \text{Length of } \overline{A'B'} = \sqrt{(6 - 4)^2 + (6 - 4)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
   \]

The final answer is \(\boxed{2\sqrt{2}}\).