Let $a_1,a_2,a_3,\dots,a_6$ be an arithmetic sequence with common difference $3$. Suppose that $a_1$, $a_3$, and $a_6$ also form a geometric sequence. Compute $a_1$.
1. Let \( a = a_1 \). Since \( a_1, a_2, a_3, \dots, a_6 \) is an arithmetic sequence with common difference \( 3 \), we can express the terms as follows:
   \[
   a_1 = a, \quad a_2 = a + 3, \quad a_3 = a + 6, \quad a_4 = a + 9, \quad a_5 = a + 12, \quad a_6 = a + 15
   \]

2. Given that \( a_1, a_3, a_6 \) form a geometric sequence, there exists a common ratio \( r \) such that:
   \[
   a_3 = a r \quad \text{and} \quad a_6 = a r^2
   \]

3. Substituting the values of \( a_3 \) and \( a_6 \) from the arithmetic sequence:
   \[
   a + 6 = a r \quad \text{and} \quad a + 15 = a r^2
   \]

4. Isolate \( r \) in the first equation:
   \[
   r = 1 + \frac{6}{a}
   \]

5. Substitute \( r \) into the second equation:
   \[
   a + 15 = a \left(1 + \frac{6}{a}\right)^2
   \]

6. Expand and simplify the equation:
   \[
   a + 15 = a \left(1 + \frac{12}{a} + \frac{36}{a^2}\right)
   \]
   \[
   a + 15 = a + 12 + \frac{36}{a}
   \]

7. Subtract \( a \) from both sides:
   \[
   15 = 12 + \frac{36}{a}
   \]

8. Isolate \( \frac{36}{a} \):
   \[
   3 = \frac{36}{a}
   \]

9. Solve for \( a \):
   \[
   a = \frac{36}{3} = 12
   \]

The final answer is \( \boxed{12} \).