Answer is B.
Here, variable $x$ of func1 points to address of variable $y.$
and variables $y$ and $z$ of func1 points to address of variable $x.$
Therefore, $y=y+4 \implies y=10+4 = 14$
and $z=x+y+z \implies z=14+14+3=31$
$z$ will be stored back in $x.$ Hence, $x=31$ and $y$ will remain as it is. $(y=3)$
Answer is $31, 3$