Identify the equations of lines through origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle $\frac{\pi}{3}$.

A. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$
B. $\frac{x}{1}=\frac{y}{-2}=\frac{z}{1}$
C. $\frac{x}{-1}=\frac{y}{-1}=\frac{z}{-2}$
D. $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$

Choose the correct answer from the options given below:

A and D only

The correct answer is Option (3) - A and D only

so $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=λ$

$x=2λ+3,y=λ+3,z=λ$ (Point of intersection)

DR of current line → (2, 1, 1) → $\vec a$

DR line through → $(2λ+3, λ+3, λ)$ → $\vec b$

$\vec a.\vec b=ab\cos \frac{π}{3}$

$\vec a.\vec b=\frac{ab}{2}$

$=2(4λ+6+λ+3+λ)=\sqrt{6}\sqrt{4λ^2+12λ+9+λ^2+6λ+9}$

$2(6λ+9)=\sqrt{6}\sqrt{6λ^2+18λ+18}$

$(2λ+3)=\sqrt{λ^2+3λ+3}$ squaring both side

$4λ^2+12λ+9=λ^2+3λ+3$

$λ^2+3λ+2=0$

$λ=-1,-2$

DR of $\vec b$ = (1, 2, -1) OR (-1, 1, -2)

lines → $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$