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Question

Find the equation of planes which passes through the intersection of the lines x+3y+6=0 and 3xy4z=0 and their perpendicular distance from the origin is 1.

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Solution

Equation of the line passing through the intersection of the given plane is given by
(x+3y+6)+k(3yy4z)=0
(1+3k)x+(3k)y4kz+6=0
If the perpendicular to it from (0,0,0) is p, then
p=∣ ∣ ∣0+00+6(1+3k)2+(3k)2+(4k)2∣ ∣ ∣
36=(1+3k)2+(3k)2+16k2
Solving k=±1
The required planes are
(x+3y+6)±(3xy4z)=0
4x+2y4z+6=0
and 2x+4y+4z+6=0

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