Step 1: Write the three power quantities of an AC circuit using their standard formulas. Apparent power \(S = VI\), real power \(P = VI\cos\phi\), reactive power \(Q = VI\sin\phi\), where \(\phi\) is the angle between voltage and current.
Step 2: From \(P = VI\cos\phi\), the unit of P works out to volt times ampere times a pure number, which is the watt, so kilowatt (kW) is the correct unit for real power. This backs up statement (A).
Step 3: From \(S = VI\), the unit of S is also volt times ampere, so kVA and not kV. A kilovolt by itself only measures potential difference and carries no current information, so it cannot be the unit of a power term. Statement (B) fails this basic check.
Step 4: Over one complete cycle, the average value of the reactive component of power is zero, it only shuttles energy into and out of the magnetic field of the load and back to the source. No net mechanical output comes from it, matching statement (C).
Step 5: A running motor behaves like an RL type load, with both a resistive part doing work and causing losses, and an inductive part building the rotating field. So it must draw both P and Q from the supply at once, confirming statement (D).
Step 6: With (A), (C) and (D) locked in as correct and (B) ruled out for its wrong unit, the option that keeps these three together, alongside (E), is the correct combination.
\[\boxed{\text{(A), (C), (D) and (E) only}}\]