1. **Problem:** Show that $\gcd(3a + 5, 7a + 12) = 1$ for $a > 0$ and $a \in \mathbb{Z}$.\n\n2. **Formula and rules:** The greatest common divisor (gcd) of two integers can be found using the Euclidean algorithm, which states that $\gcd(m, n) = \gcd(n, m \bmod n)$. If the gcd is 1, the two numbers are coprime.\n\n3. **Step-by-step solution:**\n- Compute $\gcd(3a + 5, 7a + 12)$ using the Euclidean algorithm.\n- Calculate the remainder when $7a + 12$ is divided by $3a + 5$: $$7a + 12 - 2(3a + 5) = 7a + 12 - 6a - 10 = a + 2.$$\n- So, $$\gcd(3a + 5, 7a + 12) = \gcd(3a + 5, a + 2).$$\n- Next, find $$\gcd(3a + 5, a + 2) = \gcd(a + 2, (3a + 5) - 3(a + 2)) = \gcd(a + 2, 3a + 5 - 3a - 6) = \gcd(a + 2, -1).$$\n- Since $\gcd(x, -1) = 1$ for any integer $x$, we have $$\gcd(3a + 5, 7a + 12) = 1.$$\n\n4. **Explanation:** The Euclidean algorithm reduces the gcd problem stepwise until it reaches 1, proving the two expressions are coprime for all positive integers $a$.\n\n**Final answer:** $$\boxed{1}$$