Let the image of the point (1,0,7) in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ be the point  $(\alpha ,\beta ,\gamma )$.Then which one of the following point lies on the line passing the through $(\alpha ,\beta ,\gamma )$ and making angles  $\frac{2\pi }{3}and\frac{3\pi }{4}$ with y – axis and z – axis respectively and an acute angle with x-axis ?

Given line  $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=r$
$$\begin{gathered} \text{Q(r,2r+1,3r+2)} \\ \begin{array}{l}\text{d.r's\hspace{0.17em}of\hspace{0.17em}PQ:(r-1,2r+1,3r-5)}\\ {\text{a}}_1{a}_2+b_1{b}_2+c_1{c}_2=0\Rightarrow r-1+4r+2+9r-15=0\\ 14r=14,r=1\end{array} \end{gathered}$$
$Q(1,3,5)$  let the feet of the  ${\perp }^{ler}(1,0,7)$
$R(\alpha ,\beta ,\gamma )$  be the image of p w.r.t the line
$$\begin{gathered} P+R=2Q \\ \begin{array}{l}R=2Q-P\\ (\alpha ,\beta ,r)=(2,6,10)-(1,0,7)\\ =(1,6,3)\end{array} \\ \begin{array}{l}{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\\ {\cos }^2\alpha +\frac{1}{4}+\frac{1}{2}=1\\ {\cos }^2\alpha =1-\frac{3}{4}=\frac{1}{4}\\ \cos \alpha =\frac{1}{2}\\ \alpha =\frac{\pi }{3}\end{array} \end{gathered}$$
The required line  $\frac{x-1}{\frac{1}{2}}=\frac{y-6}{\frac{-1}{2}}=\frac{z-3}{-\frac{1}{\sqrt{2}}}$
$(3,4,3-2\sqrt{2})$   lies on the line.