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The circuit is a transimpedance amplifier. It converts a constant current into a constant voltage. In this instance, with a feedback resistor of 10k, 1mA of current in, will give -10V out, because the input will stay at 0V and 1mA of curenet will give a drop of 10V across the resistor.
https://en.wikipedia.org/wiki/Transimpedance_amplifier
By biasing the op-amp to mid supply level Vcc/2, won't the input signals see the mid Vcc supply rather that pure 0V.
And if the input signals see that would there be a chance of crosstalk between them?
Yes, the (-) input will be sitting at Vcc/2 with no signal present if the (+) input is biased to Vcc/2 - ensuring both inputs are at the same voltage is the main function of an op-amp, after all - but why would that change the amount of crosstalk between mixer inputs compared to biasing the (+) input at 0V? The answer is it that it doesn't.
When the input resistors are the same value as the feedback resistor in an inverting mixer biased for single-supply operation the output of the op-amp will be sitting at Vcc/2 with no signal present, right? If you apply a 2Vpp sine wave to one of the mixer inputs then the output will go up to a peak value of (Vcc/2)+1V and a trough value of (Vcc/2)-1V; still a 2Vpp signal, except centered around Vcc/2 instead of 0V, and inverted, of course. Remember, the (+) input in a single-supply inverting amplifier is biased at a fixed voltage, so the output of the op-amp has to generate a mirror image of whatever signal is applied to the (-) input to cancel out that signal, leaving just the fixed bias voltage; this cancellation of the input signal by the op-amp output is why this configuration is said to act like a virtual earth, and it is also why the input impedance to the (-) input in this configuration is that of the series input resistor.
Since the junction of all the input resistors and the single feedback resistor is essentially earthed, every input is effectively isolated from the rest because no signal can make it past the (virtual) earth, regardless of whether that junction is at 0V or Vcc/2 or whatever.
This is kind of a tricky concept to understand, but once you get it it will make perfect sense.