Calculating Rout of a bjt while using correct RL
The load is clearly just R7.  The parallel combination of R3 and R7, is what I would call Rtc, the AC Thevenin equivalent resistance seen looking out of the collector which, in this case, is 18k.  I think it was a poor choice to label that equivalent resistance R_L.

You calculate Rout to the left of R7, i.e., do not ever, ever include the load in the output resistance calculation.

Finally, the small signal resistance looking into the collector is in parallel with R3 so, unless you're ignoring the Early effect, i.e., set r_o to infinity, the AC output resistance is not just R3.

I highly recommend consulting the late Professor Leach's ECE3050 notes:

http://users.ece.gatech.edu/mleach/ece3050/

UPDATE:

I should, I think, provide a justification for my answer above so here it is.  One of the parameters of a voltage amplifier is the open circuit voltage gain which is defined as the voltage gain when the load is an open circuit.

Now, when the load is just R7, the open circuit voltage gain is easy to compute and to measure in practice.  Just remove R7.  The circuit still functions as a voltage amplifier and the gain can be calculated or measured.

But, if R3 is part of the load, to open circuit the load, R3 must be removed!  In theory and in practice, the circuit stops functioning altogether as an amplifier.