A PIN diode is an excellent microwave switch but a poor mixer and a poor detector — so the things it cannot do are ii and iii, option 2.

The structure explains everything. A PIN diode has a thick intrinsic layer between its p and n regions. Under forward bias this layer floods with stored charge, and that charge takes a long time — the carrier lifetime, typically hundreds of nanoseconds to microseconds — to recombine.

Why the switch works so well. Above roughly

\(f\gt\dfrac{1}{2\pi\tau}\)

the stored charge cannot follow the RF cycle at all, so the diode presents a constant resistance rather than rectifying. That resistance is set by the DC bias current alone — a few tenths of an ohm when forward biased, several kilohms when reverse biased. The device therefore behaves as a current-controlled linear resistor at microwave frequencies: an ideal switch or variable attenuator, and one that introduces no distortion because it is linear to the RF signal.

Why mixing and detection fail — for the very same reason. Both operations require non-linearity at the signal frequency: a mixer must multiply two signals to produce sum and difference frequencies, and a detector must rectify. The PIN diode's slow charge storage is exactly what removes that non-linearity. The property that makes it a superb switch is the property that disqualifies it for the other two roles.

What is used instead : the Schottky barrier diode, a majority-carrier device with essentially no stored charge, which switches in picoseconds and so remains strongly non-linear well into the millimetre-wave region. Schottky diodes are the standard microwave mixer and detector elements; PIN diodes handle switching, limiting and attenuation.

The answer is flagged because PIN photodiodes are the standard optical detectors, which can make statement iii look wrong; but in the microwave context of this question, detection means RF rectification, which the PIN diode cannot perform.

Hence, a PIN diode cannot be used as a microwave mixer or detector.