If $$\frac{4x-3}{x}+\frac{4y-3}{y}+\frac{4z-3}{z}=0$$, then the value of $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ is
Given $$\frac{4x - 3}{x} + \frac{4y -3}{y} + \frac{4z-3}{z}=0$$
or $$12+(-3) ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} )$$=0
HenceÂ
 $$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$$=4
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