A) (3, 5,-2)
B) (-3,5,2)
C) (3,-5,2)
D) (3,5,2)
Correct Answer: B
Solution :
We know that, the image of the point \[({{x}_{1}},{{y}_{1}},{{z}_{1}})\]in the plane \[ax+by+cz+d=Q\]is given by \[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\] \[=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] So, the image of the point P (1, 3, 4) in the plane \[2x-y+z+3=0\]is given by \[\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\frac{-2(2-3+4+3)}{4+1+1}\] \[\Rightarrow \]\[\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2\] \[\Rightarrow \]\[x=-3,y=5,z=2\] Hence, the required point is (- 3, 5, 2).
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