A) \[\left( \frac{26}{7},\,\,\frac{15}{7},\,\,\frac{17}{7} \right)\]
B) \[\left( \frac{26}{7},\,\,\frac{-15}{7},\,\,\frac{17}{7} \right)\]
C) \[\left( \frac{15}{7},\,\,\frac{26}{7},\,\,\frac{-17}{7} \right)\]
D) \[\left( \frac{26}{7},\,\,\frac{17}{7},\,\,\frac{-15}{7} \right)\]
Correct Answer: B
Solution :
By image formula, \[\frac{{{x}_{2}}-{{x}_{1}}}{a}=\frac{{{y}_{2}}-{{y}_{1}}}{b}=\frac{{{z}_{2}}-{{z}_{1}}}{c}\] \[=-\frac{2(a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] \[\Rightarrow \] \[\frac{{{x}_{2}}-2}{3}=\frac{{{y}_{2}}+1}{-2}\,=\frac{{{z}_{2}}-3}{-1}\] \[=-2\frac{(6+2-3-9)}{9+4+1}\] \[\Rightarrow \] \[\frac{{{x}_{2}}-2}{3}=\frac{{{y}_{2}}+1}{-2}=\frac{{{z}_{2}}-3}{-1}=\frac{-2(-4)}{14}\] \[\therefore \] \[{{x}_{2}}=\frac{26}{7},\,\,{{y}_{2}}=\frac{-15}{7},\,\,{{z}_{2}}=\frac{17}{7}\]
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