A) (3, 2, 0)
B) (3, 4, 0)
C) (0, 0, 3)
D) (2, 3, 0)
Correct Answer: A
Solution :
[a] We know that z-coordinate of every point on xy-plane is zero. So, let P(x, y, 0) be a point in xy-plane such that \[PA=PB=PC.\] Now \[PA=PB\Rightarrow P{{A}^{2}}=P{{B}^{2}}\] \[\Rightarrow {{(x-2)}^{2}}+{{(y-0)}^{2}}+{{(0-3)}^{2}}\] \[={{(x-0)}^{2}}+{{(y-3)}^{2}}+{{(0-2)}^{2}}\] \[\Rightarrow 4x-6y=0\] or \[2x-3y=0...(1)\] \[PB=PC\Rightarrow P{{B}^{2}}=P{{C}^{2}}\] \[\Rightarrow {{(x-0)}^{2}}+{{(y-3)}^{2}}+{{(0-2)}^{2}}\] \[={{(x-0)}^{2}}+{{(y-0)}^{2}}+{{(0-1)}^{2}}\] \[\Rightarrow -6y+12=0\Rightarrow y=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\] Putting y = 2 in (1), we obtain x = 3. Hence, the required point is (3, 2, 0).
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