CAT 2022 Slot 1 Paper

It is given, $\mathit{y}\left(\mathit{x}+\mathit{z}\right)=19$
y cannot be 19.
If $\mathit{y}=19,\mathit{x}+\mathit{z}=1$ which is not possible when both $\mathit{x}$ and $\mathit{z}$ are natural numbers.
Therefore, $\mathit{y}=1$ and $\mathit{x}+\mathit{z}=19$
It is given, $\mathit{z}\left(\mathit{x}+\mathit{y}\right)=51$
$\mathit{z}$ can take values 3 and 17
Case 1:
If $\mathit{z}=3,\mathit{y}=1$ and $\mathit{x}=16$
$\mathit{x}\mathit{y}\mathit{z}=3\times 1\times 16=48$
Case 2:
If $\mathit{z}=17,\mathit{y}=1$ and $\mathit{x}=2$
$\mathit{x}\mathit{y}\mathit{z}=17\times 1\times 2=34$
Minimum value $\mathit{x}\mathit{y}\mathit{z}$ can take is $34$