Answer: C

Small Hint:

Let $x$ be the return time; the still-air time is then $x+9$

Big Hint:

The plane’s still-air speed is the average of its against-wind and with-wind ground speeds

Solution:

Let the distance be $d$ and the return time be $x$ minutes. The against-wind and with-wind speeds are $\frac{d}{84}$ and $\frac{d}{x}.$ Their average is the still-air speed $\frac{d}{x+9}.$ Therefore $$\frac{2}{x+9}=\frac1{84}+\frac1x.$$ Clearing denominators gives $$\begin{gathered} 168x=(x+9)(x+84),\\ x^2-75x+756=0,\\ (x-63)(x-12)=0. \end{gathered}$$ Both positive values are consistent with the stated conditions, so the two return times are $63$ and $12$ minutes.

Therefore, the correct answer is C.