A) (1, 2, 3)
B) (2, 3, 1)
C) (3, 2, 1)
D) (2, 1, 3)
Correct Answer: C
Solution :
We know that image\[(x,\,\,y,\,\,z)\]of a point \[({{x}_{1}},\,\,{{y}_{1}},\,\,{{z}_{1}})\]in a plane\[ax+by+cz+d=0\]is \[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\] \[=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] Here, point is\[(3,\,\,2,\,\,1)\]and plane is\[2x-y+3z=7\]. \[\therefore \] \[\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{3}\] \[=\frac{-2[2(3)\,-(2)+3(1)-7]}{{{2}^{2}}={{1}^{2}}+{{3}^{2}}}\] \[\Rightarrow \,\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{3}=-2(0)\] \[\Rightarrow \,\,x=3,\,y=2,\,z=1\]
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