A) \[\left( -\frac{17}{3},\,\,-\frac{19}{3},\,\,4 \right)\]
B) (15, 11, 4)
C) \[\left( -\frac{17}{3},\,\,-\frac{19}{3},\,\,1 \right)\]
D) None of these
Correct Answer: D
Solution :
If \[\left( \alpha , \beta , \gamma \right)\] be the image, then mid-point of \[\left( \alpha , \beta , \gamma \right)\] and (-1, 3, 4) must lie on \[x\text{ }-2y=0\] \[\therefore \,\,\,\frac{\alpha -1}{2}-2\,\left( \frac{\beta +3}{2} \right)=0\] \[\therefore \,\,\alpha -1-2\beta -6=\,\,0\,\,\Rightarrow \,\, \alpha \,-2\beta =7\] ... (1) Also line joining \[\left( \alpha , \beta , \gamma \right)\] and (-1, 3, 4) should be parallel to the normal of the plane \[x-2y=\text{ }0\] \[\therefore \,\,\,\frac{\alpha +1}{1}=\frac{\beta -3}{-\,2}=\frac{\gamma -4}{0}=\lambda \] \[\Rightarrow \,\,\,\alpha =\lambda -1,\,\,\beta =-\,2\lambda +3,\,\,\,\gamma =4\] ... (2) From (1) and (2) \[\alpha =\frac{9}{5},\,\,\,\beta =-\frac{13}{5},\,\,\gamma =4\] None of the option matches.
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