Which of the following sets could NOT be the lengths of the external diagonals of a right rectangular prism [a "box"]? (An [i]external diagonal[/i] is a diagonal of one of the rectangular faces of the box.)

$ \textbf{(A)}\ \{4, 5, 6\} \qquad\textbf{(B)}\ \{4, 5, 7\} \qquad\textbf{(C)}\ \{4, 6, 7\} \qquad\textbf{(D)}\ \{5, 6, 7\} \qquad\textbf{(E)}\ \{5, 7, 8\} $
1. Let the sides of the box have lengths \( x \), \( y \), and \( z \), with \( x < y < z \). The lengths of the external diagonals of the faces of the box are given by:
   \[
   \sqrt{x^2 + y^2}, \quad \sqrt{x^2 + z^2}, \quad \sqrt{y^2 + z^2}
   \]
2. We need to determine which set of lengths cannot be the lengths of these diagonals. Let's consider the set \(\{4, 5, 7\}\).

3. Suppose the lengths of the external diagonals are \(\{4, 5, 7\}\). Then we have the following equations:
   \[
   x^2 + y^2 = 16
   \]
   \[
   x^2 + z^2 = 25
   \]
   \[
   y^2 + z^2 = 49
   \]

4. Adding these three equations, we get:
   \[
   (x^2 + y^2) + (x^2 + z^2) + (y^2 + z^2) = 16 + 25 + 49
   \]
   \[
   2x^2 + 2y^2 + 2z^2 = 90
   \]
   Dividing by 2, we obtain:
   \[
   x^2 + y^2 + z^2 = 45
   \]

5. Now, subtract \( y^2 + z^2 = 49 \) from \( x^2 + y^2 + z^2 = 45 \):
   \[
   x^2 + y^2 + z^2 - (y^2 + z^2) = 45 - 49
   \]
   \[
   x^2 = -4
   \]
   This is impossible since \( x^2 \) cannot be negative.

6. Therefore, the set \(\{4, 5, 7\}\) cannot be the lengths of the external diagonals of a right rectangular prism.

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