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