Let $ s_1 = x + y + z $ and $ s_2 = xy + xz + yz $. Then \begin{align*} s_1 s_2 &= (x + y + z)(xy + xz + yz) \\ &= x^2 y + xy^2 + x^2 z + xz^2 + y^2 z + yz^2 + 3xyz \\ &= 12 + 3 \cdot 4 = 24.\end{align*}Also, \begin{align*} s_1^3 &= (x + y + z)^3 \\ &= (x^3 + y^3 + z^3) + 3(x^2 y + xy^2 + x^2 z + xz^2 + y^2 z + yz^2) + 6xyz \\ &= 4 + 3 \cdot 12 + 6 \cdot 4 = 64, \end{align*}so $ s_1 = 4 $. Hence, $ s_2 = \frac{24}{s_1} = \boxed{6} $.