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Explanation:

The given expression may be represented as

$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}$

We know that, A.M. ≥ G.M.

$\therefore \frac{\left({\displaystyle \frac{x}{y}}+{\displaystyle \frac{x}{z}}+{\displaystyle \frac{y}{x}}+{\displaystyle \frac{y}{z}}+{\displaystyle \frac{z}{x}}+{\displaystyle \frac{z}{y}}\right)}{6}\ge \sqrt[6]{\frac{x}{y}\times \frac{x}{z}\times \frac{y}{x}\times \frac{y}{z}\times \frac{z}{x}\times \frac{z}{y}}$

$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\ge 6$

∴ The given expression will have a minimum value of 6.

But x, y and z are distinct, so the value will always be greater than 6.

Hence, option (c).