If the image of the point $(4,4,3)$ in the line $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-1}{3}$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+\gamma$ is equal to:
Step-by-Step Solution
Key Concept: Find the foot of perpendicular $R$ from the point to the line, then compute the image as $2R -$ (original point).
Foot of perpendicular from $P(4,4,3)$: general point $R=(2r+1,r+2,3r+1)$.
$\overrightarrow{PR}\cdot(2,1,3)=0$: $(2r-3)\cdot2+(r-2)\cdot1+(3r-2)\cdot3=0$
$4r-6+r-2+9r-6=0 \Rightarrow 14r=14 \Rightarrow r=1$.
$R=(3,3,4)$.
Image $Q=2R-P=(6-4,6-4,8-3)=(2,2,5)=(\alpha,\beta,\gamma)$.
$\alpha+\beta+\gamma=2+2+5=9$.
Correct Answer: 1