A) \[(1,2,3)\]
B) \[(-3,2,0)\]
C) \[(3,-2,0)\]
D) \[(3,2,0)\]
E) \[(3,2,1)\]
Correct Answer: D
Solution :
Let the point in\[xy-\]plane be\[p({{x}_{1}},{{y}_{1}},0)\]. Let the given points are A (2, 0, 3), B(0, 3, 2) and C (0, 0,1). According to the given condition \[A{{P}^{2}}=B{{P}^{2}}=C{{P}^{2}}\] \[\therefore \] \[{{({{x}_{1}}-2)}^{2}}+y_{1}^{2}+9\] \[=x_{1}^{2}+{{({{y}_{1}}-3)}^{2}}+4\] \[=x_{1}^{2}+y_{1}^{2}+1\] From 1st and Und terms, \[x_{1}^{2}+4-4{{x}_{1}}+y_{1}^{2}+9\] \[=x_{1}^{2}+y_{1}^{2}-6{{y}_{1}}+9+4\] \[\Rightarrow \] \[4{{x}_{1}}-6{{y}_{1}}=0\] ...(i) From IInd and IIIrd terms, \[x_{1}^{2}+y_{1}^{2}+9-6{{y}_{1}}+4=x_{1}^{2}+y_{1}^{2}+1\] \[\Rightarrow \] \[6{{y}_{1}}=12\Rightarrow {{y}_{1}}=2\] On putting the value of\[{{y}_{1}}\]in Eq. (i) we get\[{{x}_{1}}=3\] Hence, required point is (3, 2, 0).You need to login to perform this action.
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