**Solutions:**

**(i) 12, 15 and 21**

Writing the product of prime factors for all the three numbers, we get,

12 = 2 × 2 × 3

15 = 5 × 3

21 = 7 × 3

Therefore,

HCF(12, 15, 21) = 3

LCM(12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420

**(ii) 17, 23 and 29**

Writing the product of prime factors for all the three numbers, we get,

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

Therefore,

HCF(17, 23, 29) = 1

LCM(17, 23, 29) = 17 × 23 × 29 = 11339

**(iii) 8, 9 and 25**

Writing the product of prime factors for all the three numbers, we get,

8 = 2 × 2 × 2 × 1

9 = 3 × 3 × 1

25 = 5 × 5 × 1

Therefore,

HCF(8, 9, 25) = 1

LCM(8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800