In the given circuit the current through Zener Diode is close to:

Test potential across the combination parallel to the Zener diode assuming it is disconnected: $$R_{\text{parallel}} \le R_2 = 1500\ \Omega$$

Using potential divider formula across the $$1500\ \Omega$$ resistor alone as an upper bound:

$$V_{\text{test}} = V_s \left(\frac{R_2}{R_1 + R_2}\right)$$

$$V_{\text{test}} = 12 \left(\frac{1500}{500 + 1500}\right) = 12 \times \frac{1500}{2000} = 9\text{ V}$$

Since $$V_{\text{test}} < V_Z$$ ($$9\text{ V} < 10\text{ V}$$), the potential difference across the Zener diode branch can never reach its breakdown voltage.

The Zener diode remains non-conducting: $$I_Z = 0.0\text{ mA}$$