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Question

Consider a plane x+yz=1 and point A(1,2,3). A line L has the equation x=1+3r,y=2r and z=3+4r.

The equation of the plane containing line L and point A has the equation 


A
x3y+5=0
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B
x+3y7=0
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C
3xy1=0
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D
3x+y5=0
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Solution

The correct option is C x+3y7=0
The equation of plane containing the line L is, 
A(x1)+B(y2)+C(z3)=0,
where 3AB+4C=0   ...(1)

Eqn (1) also contains point A(1,2,3).
Hence, C=0 and 3A=B 

The equation of plane is, 
x1+3(y2)=0
x+3y7=0 

Mathematics

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