Consider the planes 3x –6y + 2z + 5 = 0 and 4x –12y + 3z = 3. The plane 67x –162y + 47z + 44 = 0 bisects the angle between the given planes which

3x –6y +2z +5 = 0 ...(i); –4x +12y–3z+3=0 ...(ii)
Bisectors are
$\frac{3x-6y+2z+5}{\sqrt{9+36+4}}=\pm \frac{-4x+12y-3z+3}{\sqrt{16+144+9}}$

The plane which bisects the angle between the planes
that contains the origin
$\begin{array}{r}13(3x-6y+2z+5)=7(-4x+12y-3z+3)\\ \Rightarrow 3\times (-4)+(-6)\left(12\right)+2\times (-3)<0\dots \dots \text{ (iii) }\end{array}$
Further, $3\times (-4)+(-6)\left(12\right)+2\times (-3)<0$
Hence, the origin lies in the acute angle.