Question

If the image of the point $ (1,\,-2,3) $ in the plane $ 2x+3y-z=7 $ is the point $ (\alpha ,\beta ,\gamma ) $ then $ \alpha +\beta +\gamma $ is equal to

• A. $ -6 $
• B. $ 10 $
• C. $ 8 $
• D. $ -4 $

Answer 1

Answer: (B)

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 $