Consider two solid spherical balls, one centered at $(0, 0, \frac{21}{2} )$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac 92$ . How many points $(x, y, z)$ with only integer coordinates (lattice points) are there in the intersection of the
balls?

$\text{(A)}\ 7 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$
1. First, we need to determine the range of \( z \) values for which the two spheres intersect. The first sphere is centered at \( (0, 0, \frac{21}{2}) \) with radius 6, and the second sphere is centered at \( (0, 0, 1) \) with radius \( \frac{9}{2} \).

2. The equation of the first sphere is:
   \[
   x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \leq 6^2
   \]
   The equation of the second sphere is:
   \[
   x^2 + y^2 + (z - 1)^2 \leq \left(\frac{9}{2}\right)^2
   \]

3. To find the range of \( z \) values, we need to consider the vertical distance between the centers of the spheres. The distance between the centers is:
   \[
   \left| \frac{21}{2} - 1 \right| = \frac{19}{2}
   \]

4. The sum of the radii of the two spheres is:
   \[
   6 + \frac{9}{2} = \frac{21}{2}
   \]

5. Since the distance between the centers is less than the sum of the radii, the spheres intersect. The intersection occurs within the range of \( z \) values where both spheres' equations are satisfied.

6. We need to find the range of \( z \) values for which the intersection occurs. The intersection will be within the range:
   \[
   \frac{21}{2} - 6 \leq z \leq \frac{21}{2} + 6
   \]
   Simplifying, we get:
   \[
   \frac{9}{2} \leq z \leq \frac{33}{2}
   \]

7. Similarly, for the second sphere:
   \[
   1 - \frac{9}{2} \leq z \leq 1 + \frac{9}{2}
   \]
   Simplifying, we get:
   \[
   -\frac{7}{2} \leq z \leq \frac{11}{2}
   \]

8. The intersection of these ranges is:
   \[
   \frac{9}{2} \leq z \leq \frac{11}{2}
   \]
   Since \( z \) must be an integer, the only possible value is \( z = 5 \).

9. Substituting \( z = 5 \) into the equations of the spheres, we get:
   \[
   x^2 + y^2 + \left(5 - \frac{21}{2}\right)^2 \leq 6^2
   \]
   Simplifying, we get:
   \[
   x^2 + y^2 + \left(-\frac{11}{2}\right)^2 \leq 36
   \]
   \[
   x^2 + y^2 + \frac{121}{4} \leq 36
   \]
   \[
   x^2 + y^2 \leq 36 - \frac{121}{4}
   \]
   \[
   x^2 + y^2 \leq \frac{144}{4} - \frac{121}{4}
   \]
   \[
   x^2 + y^2 \leq \frac{23}{4}
   \]

10. For the second sphere:
    \[
    x^2 + y^2 + (5 - 1)^2 \leq \left(\frac{9}{2}\right)^2
    \]
    Simplifying, we get:
    \[
    x^2 + y^2 + 4^2 \leq \frac{81}{4}
    \]
    \[
    x^2 + y^2 + 16 \leq \frac{81}{4}
    \]
    \[
    x^2 + y^2 \leq \frac{81}{4} - 16
    \]
    \[
    x^2 + y^2 \leq \frac{81}{4} - \frac{64}{4}
    \]
    \[
    x^2 + y^2 \leq \frac{17}{4}
    \]

11. Notice that all \((x, y)\) satisfying the second inequality also satisfy the first one. Therefore, we need to find all the lattice points that satisfy \( x^2 + y^2 \leq \frac{17}{4} \).

12. The possible integer solutions for \( x \) and \( y \) are:
    \[
    (-2, 0), (2, 0), (0, -2), (0, 2), (-1, -1), (1, -1), (-1, 1), (1, 1), (-1, 0), (1, 0), (0, -1), (0, 1), (0, 0)
    \]

13. Each of these pairs corresponds to a point \((x, y, z)\) with \( z = 5 \).

Conclusion:
There are 13 lattice points in the intersection of the two spheres.

The final answer is \(\boxed{13}\)