A) \[\left( \frac{73}{25},\frac{-6}{5},\frac{39}{25} \right)\]
B) \[\left( \frac{73}{25},\frac{6}{5},\frac{39}{25} \right)\]
C) \[(-1,-2,1)\]
D) None of the above
Correct Answer: A
Solution :
Since, the plane is \[3x-5y+4z-5=0.\] If image of point\[(1,2,-1)\]in the plane is\[({{x}_{2}},{{y}_{2}},{{z}_{2}})\]then \[\frac{{{x}_{2}}-1}{3}=\frac{{{y}_{2}}-2}{-5}=\frac{{{z}_{2}}+1}{4}=\frac{-2(3-10-4)}{9+25+16}\] \[\therefore \] \[({{x}_{2}},{{y}_{2}},{{z}_{2}})=\left( \frac{73}{25},\frac{-6}{5},\frac{39}{25} \right)\]
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