A line $l$ passing through the origin is perpendicular to the lines $l_1 : (3+t)\hat{i} + (-1+2t)\hat{j} + (4+2t)\hat{k},\; -\infty < t < \infty$ $l_2 : (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k},\; -\infty < s < \infty$ Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ is (are)

$l_1:\ \text{point } (3,-1,4),\ \text{direction } \vec{d_1} = (1,2,2)$

$l_2:\ \text{point } (3,3,2),\ \text{direction } \vec{d_2} = (2,2,1)$

$l:\ \text{through } O(0,0,0),\ l \perp l_1,\ l \perp l_2$

$\text{Find } P = l \cap l_1$

$\text{Points } Q \text{ on } l_2 \text{ with } |PQ| = \sqrt{17}$

$\text{1. } l \perp l_1, l_2 \Rightarrow \vec{d} = \vec{d_1} \times \vec{d_2}$

$\text{2. Find } P = l \cap l_1$

$\text{3. } Q \in l_2:\ |PQ| = \sqrt{17}$

d⃗ = d₁⃗ × d₂⃗ = î ĵ k̂ 1 2 2 2 2 1

$\vec{d} = (-2,\ 3,\ -2)$

$l:\ (x,y,z) = \mu(-2,\ 3,\ -2)$

$l_1:\ (3+t,\ -1+2t,\ 4+2t) = \mu(-2,\ 3,\ -2)$

$t = -1,\qquad \mu = -1$

$P = (2,\ -3,\ 2)$

$Q = (3+2s,\ 3+2s,\ 2+s)$

$\vec{PQ} = Q - P = (1+2s,\ 6+2s,\ s)$

$|\vec{PQ}|^{2} = 17$

$(1+2s)^{2} + (6+2s)^{2} + s^{2} = 17$

$37 + 28s + 9s^{2} = 17$

$9s^{2} + 28s + 20 = 0$

$\Delta = 28^{2} - 4\cdot 9\cdot 20 = 784 - 720 = 64$

$s = \dfrac{-28 \pm 8}{18} = -2,\ -\dfrac{10}{9}$

$s=-2:\ Q = (-1,\ -1,\ 0)$

$s=-\dfrac{10}{9}:\ Q = \left(\dfrac{7}{9},\ \dfrac{7}{9},\ \dfrac{8}{9}\right)$

$\boxed{\text{Answer: (2) and (4)}}$