If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda }$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:

$$\begin{gathered} a_1=(1,2,3),a_2=(2,4,5),{\overrightarrow{\mathrm{b}}}_1=2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+\lambda \widehat{\mathrm{k}},{\overrightarrow{\mathrm{b}}}_2=\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+5\widehat{\mathrm{k}} \\ \mathrm{shortest} \mathrm{distance}=\frac{{\displaystyle \left|\left({\overline{\mathrm{a}}}_2-{\overline{\mathrm{a}}}_1\right)\cdot \left({\overline{\mathrm{b}}}_1\times {\overline{\mathrm{b}}}_2\right)\right|}}{\left|{\overline{\mathrm{b}}}_1\times {\overline{\mathrm{b}}}_2\right|} \end{gathered}$$

$\begin{array}{l}{\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 2& 3& \mathrm{\lambda }\\ 1& 4& 5\end{array}\right|=\widehat{\mathrm{i}\left(15-4\mathrm{\lambda }\right)}+\widehat{\mathrm{j}}\left((\mathrm{\lambda }-10\right)+\widehat{\mathrm{k}}\left(5\right)\\ \end{array}$

$\left|{\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2\right|=\sqrt{(15-4\mathrm{\lambda }{)}^2+(\mathrm{\lambda }-10{)}^2+25}$

$\text{ S.D. }=\frac{\left|\right(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})\cdot [(15-4\lambda )\widehat{\mathrm{i}}+(\lambda -10)\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}\left]\right|}{\sqrt{(15-4\lambda {)}^2+(\lambda -10{)}^2+25}}$

$\frac{|15-4\lambda +2\lambda -20+10|}{\sqrt{(15-4\lambda {)}^2+(\lambda -10{)}^2+25}}=\frac{1}{\sqrt{3}}$

$\begin{array}{l}3(5-2\lambda {)}^2=225+16{\lambda }^2-120\lambda +{\lambda }^2+100-20\lambda +25\\ 12{\lambda }^2+75-60\lambda =17{\lambda }^2-140\lambda +350\\ 5{\lambda }^2-80\lambda +275=0\\ {\lambda }^2-16\lambda +55=0\\ (\lambda -5)(\lambda -11)=0\\ \Rightarrow \lambda =5,11\end{array}$

The sum of all possible values of $\lambda$ is: 16