A) \[(3,\,5,\,-2)\]
B) \[(-3,\,5,2)\]
C) \[(3,-\,5,2)\]
D) \[(3,\,5,2)\]
Correct Answer: B
Solution :
Let \[(x,y,z)\] be the image of the point \[P(1,3,4)\] in the plane \[2x-y+z+3=0.\]Then \[\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}\] \[=\frac{-2[2(1)-1(3)+1(4)+3]}{{{(2)}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}}\] \[\therefore \]\[\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\frac{-2(6)}{6}=-2\] \[\Rightarrow \] \[x=-3,\,y=5,\,z=2\]
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