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]
$\Rightarrow$ $\begin{array}{c}x=2k+1\\ y=3k-2\\ z=-k+3\end{array}\}$ ..(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, $\alpha +\beta +\gamma =5+4+1=10$