Step 1: Breakdown into Prime Bases $(2, 3, 5)$
Left Side (LHS):
$(2^2 \cdot 3)^{12x} \cdot (2^2)^{24x+12} \cdot 5^{2y} = 2^{24x} \cdot 3^{12x} \cdot 2^{48x+24} \cdot 5^{2y} = 2^{72x+24} \cdot 3^{12x} \cdot 5^{2y}$
Right Side (RHS):
$(2^3)^{4z} \cdot (2^2 \cdot 5)^{12x} \cdot (3^5)^{3x-6} = 2^{12z} \cdot 2^{24x} \cdot 5^{12x} \cdot 3^{15x-30} = 2^{12z+24x} \cdot 3^{15x-30} \cdot 5^{12x}$
Step 2: Compare Powers of Prime Bases
For Base 3: $12x = 15x - 30 \implies 3x = 30 \implies \mathbf{x = 10}$.
For Base 5: $2y = 12x \implies 2y = 120 \implies \mathbf{y = 60}$.
For Base 2: $72x + 24 = 12z + 24x$.
Substitute $x=10$: $720 + 24 = 12z + 240 \implies 744 - 240 = 12z \implies 504 = 12z \implies \mathbf{z = 42}$.
Step 3: Final Sum
$x + y + z = 10 + 60 + 42 = \mathbf{112}$.
