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Question

Let P be the image of the point (3,1,7) with respect to the plane x−y+z=3. Then the equation of the plane passing through P and containing the straight line
x1=y2=z1 is

A
x+y3z=0
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B
3x+z=0
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C
x4y+7z=0
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D
2xy=0
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Solution

The correct option is C x4y+7z=0
We can write the equation of the line passing though P and Q. We know the coordinates of the point Q and the direction ratios of the line PQ is same as the DCs of the normal. So we get
x31=y11=z71=k
coordinate of the mid point of PQ is (k+3,k+1,k+7)
from the equation of the plane we get k=2
therefore coordinates of P is
x=1,y=5,z=3P(1,5,3)


Equation of the plane passing through P and having DCs a, b, c is given by
a(x+1)+b(y5)+c(z3)=0
Since the line with DCs 1,2,1 is perpendicular to the normal a,b,c we have
a+2b+c=0 ...(i)
We also know that point (0, 0, 0) lies on this plane, we get
a5b3c=0 ...(ii)
Using (i) and (ii), we get the ratios as
a1=b4=c7(x+1)+4(y5)7(z3)=0x+4y7z=0x4y+7z=0

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