Consider the planes $3x-6y+2z+5=0$ and $4x-12y+3z=3$.Find the equation of the plane which bisects that angle between the given planes which is acute and which contains the origin.
The two bisectors of the planes are $\frac{3x-6y+2z+5}{\sqrt{3^2+6^2+2^2}}=\pm\frac{4x-12y+3z-3}{\sqrt{4^2+12^2+3^2}}$
$\frac{3x-6y+2z+5}{7}=\pm\frac{4x-12y+3z-3}{13}$
But i could not solve it further.Please help me.