Question:

Given: Three sets of integers.
To Find: HCF and LCM of each set using prime factorisation.

(i) Numbers: 12, 15, 21
Step 1: Write prime factorisation.
$$12 = 2^{2}\times 3, \quad 15 = 3\times 5, \quad 21 = 3\times 7.$$
Step 2: Identify common prime factors. Only \(3\) is common to all three numbers.
Step 3: HCF = product of lowest powers of common primes = \(3^{1}=3\).
Step 4: For LCM, take the highest power of each prime appearing in any number: \(2^{2}, 3^{1}, 5^{1}, 7^{1}\).
Step 5: LCM = \(2^{2}\times 3\times 5\times 7 = 4\times 3\times 5\times 7 = 420\).

(ii) Numbers: 17, 23, 29
Step 1: Prime factorisation (each number is prime).
$$17 = 17, \quad 23 = 23, \quad 29 = 29.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of the highest powers (each appears to power 1).
$$\text{LCM}=17\times 23\times 29 = 11339.$$

(iii) Numbers: 8, 9, 25
Step 1: Prime factorisation.
$$8 = 2^{3}, \quad 9 = 3^{2}, \quad 25 = 5^{2}.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of highest powers: \(2^{3}, 3^{2}, 5^{2}\).
$$\text{LCM}=2^{3}\times 3^{2}\times 5^{2}=8\times 9\times 25=1800.$$

Conclusion: The HCF and LCM for each set are obtained as shown above.