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Consider the planes $3x-6y+2z+5=0$ and $4x-12y +3z=3$ . The plane $67x-16xy+4xz+44=0$ bisects the angle between the given planes which

$\begin{array}{1 1} Contains\;origin \\ is\;acuts \\ is\;obture \\ none\;if\;these \end{array} $

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Solution :
$\large\frac{4x-6y+2z+5}{\sqrt{9+36+4}}= \frac{-4x+12y-32+3}{\sqrt{16+144+9}}$
Bisects the angle between the plants the conditions
=> $ 67x -162y+47z+44=0$
$3(-4)+(-6)(12)+2(-3) <0$
Hence the origin lies in acute angle
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