
Shortcut Trick
LCM(A, B, C) = 1080 = 23 × 33 × 5.
Since HCF(A, B) = 12, HCF(B, C) = 18, and HCF(A, C) = 6, the prime factors must be distributed such that x, y, and z are pairwise coprime.
Testing x = 2, y = 3, z = 5 (all pairwise coprime):
A = 12 × 2 = 24, B = 36 × 3 = 108, C = 18 × 5 = 90.
Check: LCM(24, 108, 90) = 1080 and HCFs are correct. Sum = 24 + 108 + 90 = 222.
∴ The correct answer is 222.

Alternate Method
Given:
A = 12x, B = 36y, C = 18z
HCF(A, B) = 12, HCF(B, C) = 18, HCF(A, C) = 6
LCM(A, B, C) = 1080 = 23 × 33 × 5
Formula used: LCM of numbers is the product of the highest powers of all prime factors involved.
A = 12 × x B = 36 × y C = 18 × z LCM Factor 2³ LCM Factor 3³ LCM Factor 5 Min Sum A+B+C = 222
Prime factorization of constants:
A = 22 × 31 × x
B = 22 × 32 × y
C = 21 × 32 × z
To satisfy LCM = 23 × 33 × 51:
⇒ max(2, 2, 1, v2(x), v2(y), v2(z)) = 3
⇒ max(1, 2, 2, v3(x), v3(y), v3(z)) = 3
⇒ max(v5(x), v5(y), v5(z)) = 1
Using the pairwise coprime condition and HCF constraints:
⇒ For v2 = 3, we can set x = 2 (since A already has 22, 12 × 2 = 24 has 23).
⇒ For v3 = 3, we can set y = 3 (since B already has 32, 36 × 3 = 108 has 33).
⇒ For v5 = 1, we can set z = 5 (since C is 18 × 5 = 90).
⇒ (x, y, z) = (2, 3, 5) are pairwise coprime.
⇒ A = 12 × 2 = 24
⇒ B = 36 × 3 = 108
⇒ C = 18 × 5 = 90
⇒ A + B + C = 24 + 108 + 90 = 222.
∴ The correct answer is 222.

Additional Information
LCM and HCF Relationship
For any three numbers A, B, and C, LCM(A, B, C) × HCF(A, B) × HCF(B, C) × HCF(A, C) is not necessarily equal to A × B × C × HCF(A, B, C), unlike the case for two numbers.
Pairwise Coprime Integers
A set of integers is pairwise coprime if the HCF of every possible pair in the set is 1. If HCF(x, y) = 1, HCF(y, z) = 1, and HCF(x, z) = 1, then LCM(x, y, z) = x × y × z.
Prime Power in LCM
The exponent of a prime p in the LCM of a set of numbers is the maximum of the exponents of p in the prime factorizations of those numbers.