For real numbers, if $ |x| \plus{} |y| \equal{} 13$, then $ x^2 \plus{} 7x \minus{} 3y \plus{} y^2$ cannot be 

$\textbf{(A)}\  208 \qquad\textbf{(B)}\  15\sqrt {2} \qquad\textbf{(C)}\  \frac {35}{2} \qquad\textbf{(D)}\  37 \qquad\textbf{(E)}\ \text{None}$
1. Given the equation \( |x| + |y| = 13 \), we need to determine the values that \( x^2 + 7x - 3y + y^2 \) cannot take. 

2. We start by rewriting the expression \( x^2 + 7x - 3y + y^2 \) in a more convenient form. We complete the square for both \( x \) and \( y \):
   \[
   x^2 + 7x + \left(\frac{7}{2}\right)^2 - \left(\frac{7}{2}\right)^2 + y^2 - 3y + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2
   \]
   \[
   = \left(x + \frac{7}{2}\right)^2 - \left(\frac{7}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2
   \]
   \[
   = \left(x + \frac{7}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \frac{49}{4} - \frac{9}{4}
   \]
   \[
   = \left(x + \frac{7}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - 14.5
   \]

3. Let \( k = \left(x + \frac{7}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 \). Then the expression becomes:
   \[
   k - 14.5
   \]

4. We need to find the range of \( k \). The minimum value of \( k \) occurs when the point \((x, y)\) is closest to the center \(\left(-\frac{7}{2}, \frac{3}{2}\right)\) of the circle. The maximum value of \( k \) occurs when the point \((x, y)\) is farthest from the center.

5. The minimum distance from the center \(\left(-\frac{7}{2}, \frac{3}{2}\right)\) to the line segments defined by \( |x| + |y| = 13 \) is calculated as follows:
   \[
   \text{Distance} = \sqrt{\left(-\frac{7}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{9}{4}} = \sqrt{\frac{58}{4}} = \sqrt{14.5}
   \]
   The minimum distance is \( 4\sqrt{2} \).

6. The maximum distance from the center \(\left(-\frac{7}{2}, \frac{3}{2}\right)\) to the line segments defined by \( |x| + |y| = 13 \) is:
   \[
   \text{Distance} = \sqrt{\left(13 + \frac{7}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\left(\frac{33}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{1089}{4} + \frac{9}{4}} = \sqrt{\frac{1098}{4}} = \sqrt{274.5}
   \]

7. Therefore, the range of \( k \) is:
   \[
   14.5 \leq k \leq 274.5
   \]

8. Substituting back, the range of \( x^2 + 7x - 3y + y^2 \) is:
   \[
   0 \leq k - 14.5 \leq 260
   \]

9. We now check the given options:
   - \( 208 \) is within the range.
   - \( 15\sqrt{2} \approx 21.21 \) is within the range.
   - \( \frac{35}{2} = 17.5 \) is within the range.
   - \( 37 \) is within the range.
   - None of the above options are outside the range.

10. Hence, the answer is \( \boxed{\text{None}} \).