The image of point P (x, y, z) with respect to line x/1=(y-1)/2=(z-2)/3 is P^′(1,0,7). Find the coordinates of point P.

Let foot of the perpendicular on the given line from point P be $\displaystyle \mathrm{M}(\lambda, 2 \lambda+1,3 \lambda+2)$ D. ratios of $\displaystyle \mathbf{P P}^{\prime}$ are $\displaystyle \boldsymbol{\lambda}-\mathbf{1}, \mathbf{2} \boldsymbol{\lambda}+\mathbf{1}, \mathbf{3} \boldsymbol{\lambda}-\mathbf{5}$ $$\begin{aligned} & 1(\lambda-1)+2(2 \lambda+1)+3(3 \lambda-5)=0 \\ & \Rightarrow \lambda=1 \end{aligned} $$ Coordinates of $\displaystyle \mathrm{M}(1,3,5)$ $$\begin{aligned} & \frac{x+1}{2}=1, \frac{y+0}{2}=3, \frac{z+7}{2}=5 \\ & \Rightarrow x=1, y=6, z=3 \Rightarrow \mathrm{P}(1,6,3) \end{aligned} $$