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Given that, $x,y, \text{and}\; z$ are in A.P.

So, $y-x = z-y$

$\Rightarrow \boxed{2y = x+z}$

And, $y-x > 2\;{\color{Blue}{\text{(Increasing A.P.)}}}$

Also, $x \times y \times z = 5(x+y+z)$

$\Rightarrow x \times y \times z = 5(3y) \quad [\because x+z=2y]$

$\Rightarrow \boxed{x \times z = 15}$

$ \qquad \begin{array} {cc} \underline{x} & \underline{z} \\ 1 & 15 \\ 3 & 5 \end{array}$

Now, we can calculate the value of $y.$

$\qquad \begin{array}{} \underline{x} & \underline{y} & \underline{z} \\ 1 & 8 & 15\quad {\color{Green}{\text{in A.P.}}} \\ 3 & 4 & 5 \quad {\color{Red}{\text{in A.P.}}} \end{array}$

$\therefore$ The value of $z-x = 15-1 = 14.$

Correct Answer $:\text{C}$

So, $y-x = z-y$

$\Rightarrow \boxed{2y = x+z}$

And, $y-x > 2\;{\color{Blue}{\text{(Increasing A.P.)}}}$

Also, $x \times y \times z = 5(x+y+z)$

$\Rightarrow x \times y \times z = 5(3y) \quad [\because x+z=2y]$

$\Rightarrow \boxed{x \times z = 15}$

$ \qquad \begin{array} {cc} \underline{x} & \underline{z} \\ 1 & 15 \\ 3 & 5 \end{array}$

Now, we can calculate the value of $y.$

$\qquad \begin{array}{} \underline{x} & \underline{y} & \underline{z} \\ 1 & 8 & 15\quad {\color{Green}{\text{in A.P.}}} \\ 3 & 4 & 5 \quad {\color{Red}{\text{in A.P.}}} \end{array}$

$\therefore$ The value of $z-x = 15-1 = 14.$

Correct Answer $:\text{C}$