1. **State the problem:** Find three consecutive odd integers such that the sum of the first, twice the second, and thrice the third is 34. 2. **Define variables:** Let the first odd integer be $x$. Since the integers are consecutive odd numbers, the second integer is $x+2$ and the third is $x+4$. 3. **Write the equation:** According to the problem, $$x + 2(x+2) + 3(x+4) = 34$$ 4. **Simplify the equation:** $$x + 2x + 4 + 3x + 12 = 34$$ Combine like terms: $$x + 2x + 3x + 4 + 12 = 34$$ $$6x + 16 = 34$$ 5. **Solve for $x$:** Subtract 16 from both sides: $$6x + \cancel{16} - \cancel{16} = 34 - 16$$ $$6x = 18$$ Divide both sides by 6: $$\frac{6x}{\cancel{6}} = \frac{18}{\cancel{6}}$$ $$x = 3$$ 6. **Find the integers:** First integer: $3$ Second integer: $3 + 2 = 5$ Third integer: $3 + 4 = 7$ 7. **Check the solution:** Sum = $3 + 2(5) + 3(7) = 3 + 10 + 21 = 34$ **Answer:** The three consecutive odd integers are $3$, $5$, and $7$.