For some constant $k$ the polynomial $p(x) = 3x^2 + kx + 117$ has the property that $p(1) = p(10)$. Evaluate $p(20)$.
1. Given the polynomial \( p(x) = 3x^2 + kx + 117 \) and the condition \( p(1) = p(10) \), we need to find the value of \( k \) and then evaluate \( p(20) \).

2. First, we calculate \( p(1) \):
   \[
   p(1) = 3(1)^2 + k(1) + 117 = 3 + k + 117 = k + 120
   \]

3. Next, we calculate \( p(10) \):
   \[
   p(10) = 3(10)^2 + k(10) + 117 = 3 \cdot 100 + 10k + 117 = 300 + 10k + 117 = 10k + 417
   \]

4. According to the problem, \( p(1) = p(10) \). Therefore, we set the two expressions equal to each other:
   \[
   k + 120 = 10k + 417
   \]

5. Solving for \( k \):
   \[
   k + 120 = 10k + 417
   \]
   \[
   120 - 417 = 10k - k
   \]
   \[
   -297 = 9k
   \]
   \[
   k = -\frac{297}{9} = -33
   \]

6. Now that we have \( k = -33 \), we substitute this value back into the polynomial to find \( p(20) \):
   \[
   p(20) = 3(20)^2 + (-33)(20) + 117
   \]
   \[
   p(20) = 3 \cdot 400 - 33 \cdot 20 + 117
   \]
   \[
   p(20) = 1200 - 660 + 117
   \]
   \[
   p(20) = 1200 - 660 + 117 = 657
   \]

The final answer is \(\boxed{657}\).