Let the acute angle bisector of the two planes \(x-2 y-2 z+1=0\) and \(2 x-3 y-6 z+1=0\) be the plane \(P\). Then which …

\({P}_{1}:x-2y-2z+1=0\\ {P}_{2}:2x-3y-6z+1=0\\ \left|\frac{x-2y-2z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2x-3y-6z+1}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}\right|\\ \frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}\\ \text{ Since }{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=20>0\\ \text{∴Negative sign will give  acute bisector}\\ 7x-14y-14z+7=-[6x-9y-18z+3]\\ \Rightarrow 13x-23y-32z+10=0\\ \left(-2,0,-\frac{1}{2}\right)\text{ satisfy it. }\)