Let the acute angle bisector

$P_{1}: x-2 y-2 z+1=0$

$P_{2}: 2 x-3 y-6 z+1=0$

$\left|\frac{x-2 y-2 z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2 x-3 y-6 z+1}{\sqrt{2^{2}+3^{2}+6^{2}}}\right|$

$\frac{x-2 y-2 z+1}{3}=\pm \frac{2 x-3 y-6 z+1}{7}$

Since $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=20>0$

$\therefore$ Negative sign will give acute bisector

$7 x-14 y-14 z+7=-[6 x-9 y-18 z+3]$

$\Rightarrow 13 x-23 y-32 z+10=0$

$\left(-2,0,-\frac{1}{2}\right)$ satisfy it $\therefore$ Ans (2)