A line l ⁡ passing through the origin is perpendicular to the lines l ⁡ 1  :  3 + t i…

Given 
$l_1\text{ : }\left(3+\text{t}\right)\widehat{\text{i}}+\left(-1+2\text{t}\right)\widehat{\text{j}}+\left(4+2\text{t}\right)\widehat{\text{k}}\text{, }-\infty <\text{t}<\infty$
$l_2\text{ : }\left(3+2\text{s}\right)\widehat{\text{i}}+\left(3+2\text{s}\right)\widehat{\text{j}}+\left(2+\text{s}\right)\widehat{\text{k}}\text{, }-\infty <\text{s}<\infty$
Direction ratios of line $l_1$ are $\left(1,2,2\right)$ and line $l_2$ are $\left(2,2,1\right)$

Let equation of line $l$ is
$\text{l : }\frac{\text{x}-0}{\text{a}}=\frac{\text{y}-0}{\text{b}}=\frac{\text{z}-0}{\text{c}}=\text{k}$
This line $l$ is perpendicular to given line $l_1$ and $l_2$.
Hence $\text{a}+2\text{b}+2\text{c}=0 ..\left(i\right) \left(l\perp l_1\right)$
$2\text{a}+2\text{b}+\text{c}=0 ..\left(ii\right) \left(l\perp l_2\right)$

Using equation $\left(i\right) \& \left(ii\right)$
$\frac{\text{a}}{-2}=\frac{\text{b}}{3}=\frac{\text{c}}{-2}$

Hence equation of $l$ is $\frac{\text{x}}{-2}=\frac{\text{y}}{3}=\frac{\text{z}}{-2}=\underset{{\text{for l}}_1}{\underset{\downarrow }{{\text{k}}_1}}\text{, }\underset{{\text{for l}}_2}{\underset{\downarrow }{{\text{k}}_2}}$
Now $\text{A}\left(-2{\text{k}}_1\text{, }3{\text{k}}_1\text{, }-2{\text{k}}_1\right)\text{B}\left(-2{\text{k}}_2\text{, }3{\text{k}}_2\text{, }-2{\text{k}}_2\right)$
$-2{\text{k}}_1\widehat{\text{i}}+3{\text{k}}_1\widehat{\text{j}}-2{\text{k}}_1\widehat{\text{k}}=\left(3+\text{t}\right)\widehat{\text{i}}+\left(-1+2\text{t}\right)\widehat{\text{j}}+\left(4+2\text{t}\right)\widehat{\text{k}}$
$\begin{array}{cc}3+\text{t}=-2{\text{k}}_1& \dots \left(iii\right)\\ -1+2\text{t}=3{\text{k}}_1& \dots \left(iv\right)\\ 4+2\text{t}=-2{\text{k}}_1& \dots \left(v\right)\end{array}$
Using equation $\left(iv\right) \& \left(v\right)$ $-5=5{\text{k}}_1\Rightarrow {\text{k}}_1=-1\Rightarrow \text{A}\left(2\text{,}-3\text{,}2\right)$
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 $\sqrt{{\left(1+2\text{S}\right)}^2+{\left(6+2\text{S}\right)}^2+{\left(\text{S}\right)}^2}=\sqrt{17}$
$\Rightarrow 9{\text{S}}^2+28\text{S}+37=17$
$\Rightarrow 9{\text{S}}^2+28\text{S}+20=0$
$\Rightarrow 9{\text{S}}^2+18\text{S}+10\text{S}+20=0$
$\Rightarrow 9\text{S}\left(\text{S}+2\right)+10\left(\text{S}+2\right)=0$
$\text{⇒S}=-2\text{,}-10/9$
Using equation $\left(vi\right)$

The required points are $\left(-1,-1,0\right) \& \left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right)$