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+y−3z=0
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B
3x+z=0
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C
x−4y+7z=0
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D
2x−y=0
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Solution
The correct option is Cx−4y+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 x−31=y−1−1=z−71=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(y−5)+c(z−3)=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 a−5b−3c=0...(ii)
Using (i) and (ii), we get the ratios as a−1=b4=c−7−(x+1)+4(y−5)−7(z−3)=0−x+4y−7z=0⇒x−4y+7z=0