Find the equation of the plane which bisects the pair of planes $2x-3y+6z+2=0$ and $2x+y-2z=4$ at acute angles. 
Find the equation of the plane which bisects the pair of planes $2x-3y+6z+2=0$ and $2x+y-2z=4$ at acute angles.

My try:
The normal to the plane  $2x-3y+6z+2=0$ has direction ratios $2,-3,6$ and the normal to the plane $2x+y-2z=4$ has direction ratios $2,1,-2$.
Hence the angle between them is $\cos \theta=\vert\dfrac{n_1.n_2}{|n_1||n_2|}\vert=\dfrac{11}{45}$.
How to find the equation of the required plane from here?
Please help.
 A: Let $(x,y,z)$ be a point in the needed plane. 
Thus, $$\frac{|2x-3y+6z+2|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|2x+y-2z-4|}{\sqrt{2^2+1^2+(-2)^2}},$$
which gives equations of two our planes.
I got
$$x+2y-4z-\frac{17}{4}=0;$$
$$10x-y+2z-11=0.$$
Now, take $A(1,1,1)$, which is placed on the second plane.
Let $AB$ be a perpendicular from $A$ to the plane $2x+y-2z-4=0$ (easy to see that $AB=1$) and $AC$ be a perpendicular from $A$ to the intersection line of  given planes.
If $\measuredangle ACB<45^{\circ}$ then $10x-y+2z-11=0$ is the answer.
While if $\measuredangle ACB>45^{\circ}$ then $x+2y-4z-\frac{17}{4}=0$ is the answer.
A: HINT: The vector $\|\mathbf b\|\mathbf a + \|\mathbf a\|\mathbf b$ bisects the angle between $\mathbf a$ and $\mathbf b$.
EDIT: We need to compare the angle between $\mathbf a$ and $\mathbf b$ and the angle between $\mathbf a$ and $-\mathbf b$. One will be acute and one will be obtuse, and they want us to choose the one that is acute to bisect. As it turns out, the former is a bit more than $121^\circ$ and the latter is a bit less than $59^\circ$. So we should work with the latter.  
(Note that $(2,-3,6)\cdot (2,1,-2) < 0$ whereas $(2,-3,6)\cdot (-2,-1,2)>0$.)
