$a, b, c, x, y, z ...$

Just assume that $a=b=c$

Then, $x=3, y=3 ,z=3$

Therefore, only option that corresponds to $xy+yz+zx$ is $xyz$

${Actual Method}$

$a^{x-1} = bc$

$a^x=abc$

Similarly,

$b^y=abc$

$c^z=abc$

Therefore,

$a^x=b^y$

$b=a^{x/y}$

$ab=c^z/c. (abc=c^z)$

$a^{x/y + 1}= c^{z-1}$

$a^{x/y + 1}=a^x/c$

$a^{x/y + 1}=a^x/(a^{x/z})$

Therefore,

$a^{x/y + 1} = a^{x- x/z}$

Therefore,

$x/y + 1 = x - x/z$

$xy+yz=xyz-xy$

$xy+yz+xy = xyz$