Let $ABC$ be a triangle with $AB = 9$, $BC = 10$, and $CA = 17$. Let $B'$ be the reflection of the point $B$ over the line $CA$. Let $G$ be the centroid of triangle $ABC$, and let $G'$ be the centroid of triangle $AB'C$. Determine the length of segment $GG'$.
1. **Identify the key points and segments:**
   - Given triangle \(ABC\) with \(AB = 9\), \(BC = 10\), and \(CA = 17\).
   - \(B'\) is the reflection of \(B\) over the line \(CA\).
   - \(G\) is the centroid of triangle \(ABC\).
   - \(G'\) is the centroid of triangle \(AB'C\).

2. **Determine the relationship between \(G\) and \(G'\):**
   - Let \(H\) be the intersection of \(\overline{BB'}\) and \(\overline{AC}\).
   - Let \(J\) be the intersection of \(\overline{GG'}\) and \(\overline{AC}\).
   - It is known that \(GJ = \frac{1}{3}BH\), so \(GG' = \frac{2}{3}BH\).

3. **Calculate the area of \(\triangle ABC\) using Heron's formula:**
   - The semi-perimeter \(s\) of \(\triangle ABC\) is:
     \[
     s = \frac{AB + BC + CA}{2} = \frac{9 + 10 + 17}{2} = 18
     \]
   - Using Heron's formula:
     \[
     [ABC] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18 \cdot (18-9) \cdot (18-10) \cdot (18-17)}
     \]
     \[
     [ABC] = \sqrt{18 \cdot 9 \cdot 8 \cdot 1} = \sqrt{18 \cdot 72} = \sqrt{1296} = 36
     \]

4. **Find the altitude \(BH\) of \(\triangle ABC\) to base \(AC\):**
   - The area of \(\triangle ABC\) can also be expressed as:
     \[
     [ABC] = \frac{1}{2} \cdot AC \cdot BH
     \]
   - Solving for \(BH\):
     \[
     36 = \frac{1}{2} \cdot 17 \cdot BH \implies BH = \frac{2 \cdot 36}{17} = \frac{72}{17}
     \]

5. **Calculate the length of segment \(GG'\):**
   - Since \(GG' = \frac{2}{3}BH\):
     \[
     GG' = \frac{2}{3} \cdot \frac{72}{17} = \frac{144}{51} = \frac{48}{17}
     \]

The final answer is \(\boxed{\frac{48}{17}}\).