CAT Exam 2021 Slot 2 Solved Paper

Given x, y, z are three terms in an arithmetic progression.
Considering $\mathit{x}=\mathit{a},\mathit{y}=\mathit{a}+\mathit{d}$, $\mathit{z}=\mathit{a}+2\times \mathit{d}$
Using the given equation $\mathit{x}\times \mathit{y}\times \mathit{z}=5\times \left(\mathit{x}+\mathit{y}+\mathit{z}\right)$
$\mathit{a}\times (\mathit{a}+\mathit{d}\times \left(\mathit{a}+2\times \mathit{d}\right)=5\times \left(\mathit{a}+\mathit{a}+\mathit{d}+\mathit{a}+2\times \mathit{d}\right)$
$\mathit{a}\times \left(\mathit{a}+\mathit{d}\right)\times \left(\mathit{a}+2\times \mathit{d}\right)=5\times \left(3\times \mathit{a}+3\times \mathit{d}\right)=15\times \left(\mathit{a}+\mathit{d}\right)$
$\mathit{a}\times \left(\mathit{a}+2\times \mathit{d}\right)=15$
Since all x, y, z are positive integers and $\mathit{y}-\mathit{x}>2.\mathit{a},\mathit{a}+\mathit{d},\mathit{a}+2\times \mathit{d}$ are integers.
The common difference is positive and greater than 2.
Among the different possibilities are : $\left(\mathit{a}=1,\mathit{a}+2\mathit{d}=5\right),\left(\mathit{a},=3,\mathit{a}+2\mathit{d}=5\right),\left(\mathit{a}=5,\mathit{a}+2\mathit{d}=3\right),\left(\mathit{a}=15,\mathit{a}+2\mathit{d}=1\right)$
Hence the only possible case satisfying the condition is :
$\mathit{a}=1,\mathit{a}+2\times \mathit{d}=15$
$\mathit{x}=1,\mathit{z}=15$
$\mathit{z}-\mathit{x}=14$