If $$12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20 ^{12x} \times 243^{3x-6}$$, where x , y and z are
natural numbers, then $$ x + y + z $$ equals

$$12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20 ^{12x} \times 243^{3x-6}$$

On rewriting after prime factorisation, we get,

$$\cancel{2^{24x}}\times 3^{12x} \times 2^{48x+24}\times 5^{2y} = 2^{12z}\times \cancel{2^{24x}} \times 5^{12x} \times 3^{15x-30}$$

Since LHS = RHS, the corresponding powers must be equal. We have,

Powers of $$3$$:  $$12x = 15x-30$$  or  $$x=10$$, and

Powers of $$2$$:  $$48x+24 = 12z$$  or  $$4x+2 = z$$  or  $$z = 42$$

Powers of $$5$$:  $$2y = 12x$$  or  $$y= 6x$$  or  $$y=60$$

Therefore, $$x+y+z = 10+42+60 = 112$$