A) \[-6\]
B) \[10\]
C) \[8\]
D) \[-4\]
Correct Answer: B
Solution :
Given points is \[(1,\,-2,3)\] and plane is \[2x+3y-z=7\] Equation of line passing through the point \[(1,-2,3)\] and perpendicular to the given plane is \[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-1}=k\] [say] \[\Rightarrow \] \[\left. \begin{matrix} x=2k+1 \\ y=3k-2 \\ z=-k+3 \\ \end{matrix} \right\}\] ?..(i) This is the common point for plane and line passing through the point \[(1,\,\,-2,\,\,3)\]. \[\therefore \] \[2(2k+1)+3(3k-2)-(-k+3)=7\] \[\Rightarrow \] \[4k+2+9k-6+k-3=7\] \[\Rightarrow \] \[14k-7=7\Rightarrow 14k=14\] \[\Rightarrow \] \[k=1\] Then, from Eq. (i), we get \[x=2\times 1+1=3\] \[y=3\times 1-2=1\] \[z=-1+3=2\] Given image of points \[(1,-2,3)\] is \[(\alpha ,\beta ,\gamma ).\] \[\therefore \] \[\frac{\alpha +1}{2}=3,\,\,\frac{\beta -2}{2}=1,\frac{\gamma +3}{2}=2\] [\[\because \] point \[(3,1,2)\] is the mid-point of the line joining \[(1,-2,3)\] and \[(\alpha ,\beta ,\gamma ]\] \[\Rightarrow \] \[\alpha +1=6,\,\beta -2=2,\,\gamma +3=4\] \[\Rightarrow \] \[\alpha =5,\beta =4,\gamma =1\] Now, \[\alpha +\beta +\gamma =5+4+1=10\]
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