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Question

The image of the line x13=y31=z45 in the plane 2xy+z+3=0 is the line
a) x+33=y51=z25
b) x+33=y51=z+25
c) x33+y+51=z25
d) x33=y+51=z25

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Solution

Point (1,3,4)
Solution of normal to plane is
x12=y31=z4j=k
Any point on the normal is
(2k+1,k+3,k+4) (It lies on plane)
2(k+1)(6k2)+8+k2+3=0
k=2
Point through which image passing (-3, 5, 2)
So, required equation of line
x+33=y51=z55

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