Given a rectangle $ABCD$ with a point $P$ inside it. It is known that $PA =
17, PB = 15,$ and $PC = 6.$
What is the length of $PD$?
1. Let the distances from point \( P \) to the vertices \( A, B, C, \) and \( D \) be denoted as \( PA = 17, PB = 15, \) and \( PC = 6 \). We need to find the distance \( PD \).

2. We can use the British flag theorem, which states that for any point \( P \) inside a rectangle \( ABCD \), the sum of the squares of the distances from \( P \) to two opposite corners is equal to the sum of the squares of the distances from \( P \) to the other two opposite corners. Mathematically, this is expressed as:
   \[
   PA^2 + PC^2 = PB^2 + PD^2
   \]

3. Substitute the given values into the equation:
   \[
   17^2 + 6^2 = 15^2 + PD^2
   \]

4. Calculate the squares:
   \[
   289 + 36 = 225 + PD^2
   \]

5. Simplify the equation:
   \[
   325 = 225 + PD^2
   \]

6. Isolate \( PD^2 \):
   \[
   PD^2 = 325 - 225
   \]

7. Simplify further:
   \[
   PD^2 = 100
   \]

8. Take the square root of both sides to find \( PD \):
   \[
   PD = \sqrt{100} = 10
   \]

The final answer is \( \boxed{10} \).