If $ x^{2} \plus{}y^{2} \plus{}z\equal{}15$, $ x\plus{}y\plus{}z^{2} \equal{}27$ and $ xy\plus{}yz\plus{}zx\equal{}7$, then

$\textbf{(A)}\  3\le \left|x\plus{}y\plus{}z\right|\le 4 \\ \textbf{(B)}\  5\le \left|x\plus{}y\plus{}z\right|\le 6 \\ \textbf{(C)}\  7\le \left|x\plus{}y\plus{}z\right|\le 8 \\ \textbf{(D)}\  9\le \left|x\plus{}y\plus{}z\right|\le 10 \\ \textbf{(E)}\ \text{None}$
1. Denote \( S = x + y + z \). We are given the following equations:
   \[
   x^2 + y^2 + z = 15
   \]
   \[
   x + y + z^2 = 27
   \]
   \[
   xy + yz + zx = 7
   \]

2. First, multiply the second equation by 2:
   \[
   2(x + y + z^2) = 2 \cdot 27
   \]
   \[
   2x + 2y + 2z^2 = 54
   \]

3. Add the first equation to the modified second equation:
   \[
   x^2 + y^2 + z + 2x + 2y + 2z^2 = 15 + 54
   \]
   \[
   x^2 + y^2 + 2z^2 + z + 2x + 2y = 69
   \]

4. Notice that we can rewrite the left-hand side as:
   \[
   x^2 + y^2 + 2z^2 + z + 2x + 2y = (x^2 + 2x + 1) + (y^2 + 2y + 1) + 2z^2 + z - 2
   \]
   \[
   = (x+1)^2 + (y+1)^2 + 2z^2 + z - 2
   \]

5. Now, we need to consider the third equation \( xy + yz + zx = 7 \). We can use the identity for the square of a sum:
   \[
   (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
   \]

6. Substitute the known values:
   \[
   S^2 = x^2 + y^2 + z^2 + 2 \cdot 7
   \]
   \[
   S^2 = x^2 + y^2 + z^2 + 14
   \]

7. From the first equation, we know \( x^2 + y^2 + z = 15 \). Let \( z = 15 - x^2 - y^2 \):
   \[
   S^2 = 15 + z^2 + 14
   \]
   \[
   S^2 = 29 + z^2
   \]

8. From the second equation, we know \( x + y + z^2 = 27 \). Let \( z^2 = 27 - x - y \):
   \[
   S^2 = 29 + (27 - x - y)
   \]
   \[
   S^2 = 56
   \]

9. Therefore, \( S^2 = 56 \) implies:
   \[
   S = \sqrt{56} \quad \text{or} \quad S = -\sqrt{56}
   \]
   \[
   S = 7.48 \quad \text{or} \quad S = -7.48
   \]

10. Hence, \( 7 \leq |S| \leq 8 \).

The final answer is \( \boxed{7 \leq |x + y + z| \leq 8} \)