Find the equations of the lines through the origin which intersect th…

The correct answer is Option (1) → $\frac{x}{1} = \frac{y}{2} = \frac{z}{-1}$ and $\frac{x}{-1} = \frac{y}{1} = \frac{z}{-2}$ ##

Let $P(2\lambda + 3, \lambda + 3, \lambda)$ be any point on line

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

Let the line through origin and making an angle of $\frac{\pi}{3}$ with the given line be along $OP$. Then direction ratios are proportional to $2\lambda + 3 - 0, \lambda + 3 - 0, \lambda - 0$ i.e., $2\lambda + 3, \lambda + 3, \lambda$

Also, direction ratios of the given line are proportional to $2, 1, 1$.

$∴\cos \frac{\pi}{3} = \frac{(2\lambda + 3)(2) + (\lambda + 3)(1) + (\lambda)(1)}{\sqrt{(2\lambda + 3)^2 + (\lambda + 3)^2 + (\lambda)^2} \sqrt{2^2 + 1^2 + 1^2}}$

$\Rightarrow \frac{1}{2} = \frac{6\lambda + 9}{\sqrt{6\lambda^2 + 18\lambda + 18} \sqrt{6}}$

$\Rightarrow \frac{1}{2} = \frac{3(2\lambda + 3)}{6\sqrt{\lambda^2 + 3\lambda + 3}}$

$\Rightarrow \sqrt{\lambda^2 + 3\lambda + 3} = (2\lambda + 3)$

Squaring both sides, we get

$\lambda^2 + 3\lambda + 3 = (2\lambda + 3)^2$

$\Rightarrow \lambda^2 + 3\lambda + 3 = 4\lambda^2 + 12\lambda + 9$

$\Rightarrow 3\lambda^2 + 9\lambda + 6 = 0$

$\Rightarrow \lambda^2 + 3\lambda + 2 = 0$

$\Rightarrow \lambda = -1, -2$

Therefore, the coordinates of point $P(1, 2, -1)$ or $P(-1, 1, -2)$

Hence, Equations of required lines are

$\frac{x}{1} = \frac{y}{2} = \frac{z}{-1} \text{ and } \frac{x}{-1} = \frac{y}{1} = \frac{z}{-2}$