If a line passing through the origin is perpendicular to the lines :…

Let the equation of a line passing through the origin and perpendicular to $l_1$ and $l_2$ be $\vec{r} = s \vec{b}.$

This is perpendicular the $l_1$ and $l_2$ both. So, $\vec{b}$ is parallel to the vector $\vec{b_1}×\vec{b_2}$, where $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2}= 2\hat{i} + 2\hat{j} + \hat{k}$ are vectors parallel to $l_1$ and $l_2$ respectively.

Now, 

$\vec{b_1}×\vec{b_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 2 & 2\\2 & 2 & 1\end{vmatrix}= 2\hat{i} + 3\hat{j} - 2\hat{k}$

So, the vector equation of line $ l $ is $\vec{r}= s(-2\hat{i} + 3\hat{j} - 2\hat{k})$

At the point of intersection of l i.e. $\vec{r} = s (-2 \hat{i}+3\hat{j} - 2\hat{k})$ and $l_1$

i.e. $ \vec{r} = ( 3 + t) \hat{i} + (-1 + 2t) \hat{j} + (4 + 2t) \hat{k} $, we have

$3 + t = -2s, -1 + 2t = 3s $ and $4 + 2t = -2s$

$⇒ s = -1 $ and $ t = - 1$

The position vector of the point of intersection of $l$ and $l_1$ is $2\hat{i} - 3\hat{j} + 2\hat{k}$. Let this point be P.

Let Q be a point on $l_2$ such that $PQ= \sqrt{17}$.

Let the position vector of Q be $(3 + 2t) \hat{i} + (3 + 2t) \hat{j} + ( 2 + t)\hat{k}.$

Then,

$PQ= \sqrt{17} ⇒ |\vec{PQ}| = \sqrt{17}$

$⇒ |(1+2t)\hat{i} + (6 + 2t) \hat{j} + t\hat{k}|= \sqrt{17}$

$⇒ (2t+2)^2 + (2t + 6)^2 + t^2 = 17 ⇒ 9t^2 + 28t + 20 = 0 $

$⇒ (t+2) (9t + 10) = 0 ⇒ t = -2, -\frac{10}{9}$

Thus , the position vectors of Q are $\hat{i} + \hat{j} + \hat{k} $ or $\frac{7}{9}\hat{i} +\frac{7}{9}\hat{j}+\frac{8}{9}\hat{k}.$

Consequently, coordinates of Q can be (1, 1, 1) and $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right) $