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Question

Find the equation of the bisector planes of the angles between the planes 2x−y+2z+3=0 and 3x−2y+6z+8=0.

A
5xy4z22=0
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B
23x13y+32z+26=0
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C
19xy4z+26=0
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D
none of these
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Solution

The correct options are
A 5xy4z22=0
B 23x13y+32z+26=0
Equation of the planes is 2x4y+2z+3=0 and 3x2y+6z+8=0
Then equation of the plane bisection the angles between them are
2x4y+2z+34+16+4+9=±3x2y+6z+89+4+36+64
2x4y+2z+333=±3x2y+6z+8113
5xy4z22=0 and 23x13y+32z+26=0

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