plz answer this question with explanation
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Step-by-step explanation:
Let 2^x = 3^y = 6^-z = k
=> 2^x = k
=> k^1/x = 2
similarly k^1/y = 3 and k^1/-z = 6
=>k^1/-z = 2×3
=> k^-1/z = k^1/x × k^1/y
=> k-^1/z = k^(1/x+1/y)
=>-1/z = 1/x + 1/y
=> 1/x + 1/y + 1/z = 0
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