Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane $P$. Then which of the following points lies on $P ?$

Angle bisectors of the two given planes are

$\frac{x-2y-2z+1}{\sqrt{1+4+4}}=\pm \frac{2x-3y-6z+1}{\sqrt{4+9+36}}$

$\Rightarrow x-5y+4z+4=0\text{ and }13x-23y-32z+10=0$

$\text{ Let }\theta \text{ be the angle between the planes }x-5y+4z+4=0\text{ and }x-2y-2z+1=0$

So, $\cos \theta =\frac{1+10-8}{\sqrt{1+4+4}·\sqrt{1+25+16}}=\frac{1}{\sqrt{42}}\Rightarrow \tan \theta =\sqrt{41}>1$

$\Rightarrow \theta >45°$

Then acute angle bisector is plane $P: 13x-23y-32z+10=0$

Point $\left(-2, 0, -\frac{1}{2}\right)$ lies on the the plane $P$