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Let 2x = 3y = 6-z = k So, 2 = k(1/x) , 3 = k(1/y) and 6 = k(-1/z) k(1/x) × k(1/y) = k(-1/z) [:. 2 x 3 = 6] k(1/x + 1/y) = k(-1/z) (1/x) + (1/y) = (-1/z) (1/x) + (1/y) + (1/z) = 0
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