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

<p>Let ${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$</p> <p>${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$</p> <p>$$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$$</p> <p>$\therefore$ $$\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat k$$</p> <p>$${\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat k$$</p> <p>$\therefore$ Shortest distance</p> <p>$$ = \left| {{{(15 - 4\lambda ) - 2(10 - \lambda ) + 10} \over {\sqrt {{{(15 - 4\lambda )}^2} + {{(10 - \lambda )}^2} + 25} }}} \right| = {1 \over {\sqrt 3 }}$$</p> <p>$$ \Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25$$</p> <p>$\Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0$</p> <p>$\therefore$ Sum of values of $\lambda = {{80} \over 5} = 16$</p>