The image of the point (1, 3, 4) in the plane `2x-y+z+3=0` is `(3,\ 5,\ 2)` b. `(-3,5,\ 2)` c.`(3,\ 5,\ -2)` d. `(3,\ -5,\ 2)`

Explanation

The equation of a line through the point `P(1,3,4)` and perpendicular to the plane `2 x-y+z+3=0` is

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`frac{x-1}{2}=frac{y-3}{-1}=frac{z-4}{1}=k text { (say) }`


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A general point on this line is `(2 k+1,-k+3, k+4)`.


For some value of k let `N(2 k+1,-k+3, k+4)` be a point of the line lying on the plane. Then,


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`2(2 k+1)-(-k+3)+(k+4)+3=0 Rightarrow 6 k=-6 Rightarrow k=-1 text {. }`


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