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)

$l_1: \vec{r}=(3\hat{i}-\hat{j}+4\hat{k})+t(\hat{i}+2\hat{j}+2\hat{k})$

$l_2: \vec{r}=(3\hat{i}+3\hat{j}+2\hat{k})+s(2\hat{i}+2\hat{j}+\hat{k})$

$l \perp l_1,\quad l \perp l_2,\quad l \text{ passes through the origin}$

$\text{To find: points on } l_2 \text{ at distance } \sqrt{17} \text{ from } l\cap l_1$

$\vec{a}=(1,2,2),\quad A(3,-1,4)$

$\vec{b}=(2,2,1),\quad B(3,3,2)$

$\vec{d}=\vec{a}\times\vec{b}\ \perp\ \vec{a}\ \text{and}\ \vec{b}$

$\text{Plan: } 1.\ \vec{d}=\vec{a}\times\vec{b}$

$2.\ P = l \cap l_1$

$3.\ |Q-P|=\sqrt{17},\ \ Q\in l_2$

a⃗×b⃗= î ĵ k̂ 1 2 2 2 2 1

$= \hat{i}(2-4)-\hat{j}(1-4)+\hat{k}(2-4)$

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

$l: \vec{r}=\lambda(-2,\ 3,\ -2)$

$(3+t,\ -1+2t,\ 4+2t)=(-2\lambda,\ 3\lambda,\ -2\lambda)$

$3+t=-2\lambda,\quad -1+2t=3\lambda,\quad 4+2t=-2\lambda$

$3+t=4+2t \Rightarrow t=-1$

$\lambda=-1 \Rightarrow P=(2,\ -3,\ 2)$

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

$Q-P=(1+2s,\ 6+2s,\ s)$

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

$9s^2+28s+37=17 \ \Rightarrow\ 9s^2+28s+20=0$

$s=\dfrac{-28\pm\sqrt{28^2-4(9)(20)}}{18}=\dfrac{-28\pm 8}{18}$

$s=-2 \Rightarrow Q_1=(-1,\ -1,\ 0)$

$s=-\dfrac{10}{9} \Rightarrow Q_2=\left(\dfrac{7}{9},\ \dfrac{7}{9},\ \dfrac{8}{9}\right)$

$Q_1=(-1,\ -1,\ 0)\ \Rightarrow\ \text{option (2)}$

$Q_2=\left(\dfrac{7}{9},\ \dfrac{7}{9},\ \dfrac{8}{9}\right)\ \Rightarrow\ \text{option (4)}$

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