Multiply the three equations

$(xy)(yz)(xz) = -3(12)(-4)$

$x^2 y^2 z^2 = 144$

$(xyz)^2 = 144$

$xyz = \pm 12$ → Equation (4)

Equation (4) divided by Equation (2)

$\dfrac{xyz}{yz} = \dfrac{\pm 12}{12}$

$x = \pm 1$ *answer*

Equation (4) divided by Equation (3)

$\dfrac{xyz}{xz} = \dfrac{\pm 12}{-4}$

$y = \pm 3$ *answer*

Equation (4) divided by Equation (1)

$\dfrac{xyz}{xy} = \dfrac{\pm 12}{-3}$

$z = \pm 4$ *answer*