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Question

Equation of the plane containing the line x+2y+3z4=0=2x+yz+5 and perpendicular to the plane 5x+3y6z+8=0 is

A
33x+45y+50z=41
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B
33x+45y+50z=41
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C
35x+45y+50z=41
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D
35x+45y+50z=41
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Solution

The correct option is A 33x+45y+50z=41
Equation of the plane containing the given line is
P:(x+2y+3z4)+λ(2x+yz+5)=0
P:x(1+2λ)+y(2+λ)+z(3λ)+(5λ4)=0

Since, P is perpendicular to 5x+3y6z+8=0.
5(1+2λ)+3(2+λ)6(3λ)=0λ=719

The required equation of the plane is 33x+45y+50z=41

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