A line l passing through the origin is perpendicular to the lines l...

Given 
{l }_{1} \text{:} \left(3 + \text{t}\right) \overset{^}{\text{i}} + \left(- 1 + 2 \text{t}\right) \overset{^}{\text{j}} + \left(4 + 2 \text{t}\right) \overset{^}{\text{k}} \text{,} - \infty < \text{t} < \infty
{l }_{2} \text{:} \left(3 + 2 \text{s}\right) \overset{^}{\text{i}} + \left(3 + 2 \text{s}\right) \overset{^}{\text{j}} + \left(2 + \text{s}\right) \overset{^}{\text{k}} \text{,} - \infty < \text{s} < \infty
Direction ratios of line ${l}_{1}$ are \left(1, 2, 2\right) and line ${l}_{2}$ are \left(2, 2, 1\right)$</p><p>Let equation of line$l$is<br >$\text{l :}\dfrac{\text{x} - 0}{\text{a}} = \dfrac{\text{y} - 0}{\text{b}} = \dfrac{\text{z} - 0}{\text{c}} = \text{k}$<br >This line$l$is perpendicular to given line${l}_{1}$and${l}_{2}$.<br >Hence$\text{a} + 2\text{b} + 2\text{c} = 0 . . \left(i\right)\left(l ⊥ {l}_{1}\right)$<br >$2\text{a} + 2\text{b} + \text{c} = 0 . . \left(ii\right)\left(l ⊥ {l}_{2}\right)</p><p>Using equation \left(i\right) & \left(ii\right)
$\dfrac{\text{a}}{- 2} = \dfrac{\text{b}}{3} = \dfrac{\text{c}}{- 2}$

Hence equation of $l$ is $\dfrac{\text{x}}{- 2} = \dfrac{\text{y}}{3} = \dfrac{\text{z}}{- 2} = \mathop{\mathop{{\text{k}}_{1}}_{↓}}_{{\text{for l}}_{1}}\text{,}\mathop{\mathop{{\text{k}}_{2}}_{↓}}_{{\text{for l}}_{2}}$
Now $\text{A} \left(- 2 {\text{k}}_{1} \text{,} 3 {\text{k}}_{1} \text{,} - 2 {\text{k}}_{1}\right) \text{B} \left(- 2 {\text{k}}_{2} \text{,} 3 {\text{k}}_{2} \text{,} - 2 {\text{k}}_{2}\right)$
- 2 {\text{k}}_{1} \overset{^}{\text{i}} + 3 {\text{k}}_{1} \overset{^}{\text{j}} - 2 {\text{k}}_{1} \overset{^}{\text{k}} = \left(3 + \text{t}\right) \overset{^}{\text{i}} + \left(- 1 + 2 \text{t}\right) \overset{^}{\text{j}} + \left(4 + 2 \text{t}\right) \overset{^}{\text{k}}
$\begin{cases}3 + \text{t} = - 2{\text{k}}_{1} & \ldots \left(iii\right) \\ - 1 + 2\text{t} = 3{\text{k}}_{1} & \ldots \left(iv\right) \\ 4 + 2\text{t} = - 2{\text{k}}_{1} & \ldots \left(v\right)\end{cases}$
Using equation \left(iv\right) & \left(v\right)$- 5 = 5 {\text{k}}_{1} ⇒ {\text{k}}_{1} = - 1 ⇒ \text{A} \left(2 \text{,} - 3 \text{,} 2\right)$
Let any point on ${l}_{2}\left(3 + 2\text{S}\text{,}3 + 2\text{S}\text{,}2 + \text{S}\right) . . . . \left(vi\right)$
According to question $\sqrt{{\left(1 + 2 \text{S}\right)}^{2} + {\left(6 + 2 \text{S}\right)}^{2} + {\left(\text{S}\right)}^{2}} = \sqrt{1 7}$
$⇒ 9{\text{S}}^{2} + 28\text{S} + 37 = 17$
$⇒ 9{\text{S}}^{2} + 28\text{S} + 20 = 0$
$⇒ 9{\text{S}}^{2} + 18\text{S} + 10\text{S} + 20 = 0$
$⇒ 9\text{S}\left(\text{S} + 2\right) + 10\left(\text{S} + 2\right) = 0$
$\text{⇒S} = - 2\text{,} - 10 / 9$
Using equation \left(vi\right)</p><p>The required points are \left(- 1, - 1, 0\right) & \left(\dfrac{7}{9}, \dfrac{7}{9}, \dfrac{8}{9}\right)