The given logical expression can be reduces as follows:
$\implies \bar xy\bar z+\bar xyz+xy\bar z+xyz$
Taking common
$\implies \bar xy(z+\bar z)+xy(z+\bar z)$
$\because (x+\bar x=1)$
$\implies \bar xy+xy$
taking $y$ as common we get:
$\implies y(x+\bar x)$
$\implies y$
option C is correct.