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Question

Find the equation of a plane which passes through the point (1,2,3) and which is equally inclined to the planes x2y+2z3=0 and 8x4y+z7=0.

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Solution

Equation of angle bisector of a given plane:
x2y+2z39=±(8x4y+z7)64+16+7x2y+2z33=±(8x4y+z7)93(x2y+2z3)=8x4y+z75x5z+2=0takingvesign3(x2y+2z3)=(8x4y+z7)11x10y+7z16=0Equationofaplaneis:5x5z+c1=055(3)+c1=0c1=105x5z+10=011x10y+7z+c2=01120+21+c2=0c2=1211x10y+7z12=0

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