Electrical – Why we put 10Kohm resistor in parallel with 144Kohm that exist inside USB-6008?

I think your system relies on all the photo-transistors being illuminated when no object is passing in front. With 6 photo-transistors in series and all of them being strongly illuminated, the "on" voltage of the group is 6 x 0.4V = 2.4 volts.

The 0.4V figure comes from the data sheet, \$V_{CE(SAT)}\$ = 0.4V at 20mW/sq cm illumination.

Any device switching off (due to a blockage) will cause the signal to rise to about 5V. So, the basic system you have cannot be guaranteed to generate any change in signal voltage greater than 2.6 volts p-p and this is with fully illuminated devices. Should the IR diode that does the illumination not be producing a power output of at least 20mW/sq cm at the photo-transistors, the individual "on" voltages will be somewhat greater than 0.4 volts and this might be (say) 0.8 volts.

The impact of this is that the change in signal voltage seen at the output will be 5 volts - (6 x 0.8 volts) = 0.2 volts.

I believe you need to sort out this problem first before deciding on the value of resistor. Look at the volt drop across one of the photo-transistors (not in the group but separately powered via a load resistor) and see what it is when illuminated.

I'll also add that the data sheet is not very helpful when it comes to stating what the "illuminated" collector current is - there seems to be several figures and no explanation of what conditions these different figures apply to.

Is there any chance that there is a problem with either the connections or the bridge itself, particularly on the (+) input?

My suggestion: replace the bridge with 4 resistors (match them if you wish) and see what happens.

This really is starting to look like a bad connection somewhere around the (+) input of the op-amp.