The right-hand diode network selects the highest source, +15 V. On the left, the +6 V branch receives current through 2 kΩ; the +12 V diode remains off because the junction voltage stays below 12 V. The active series path therefore spans $15-6=9$ V across $3+2=5$ kΩ, giving $I_k=9/5=1.8$ mA. The intervening left junction is $6+1.8(2)=9.6$ V, consistent with the assumed diode states.
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