Let $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$$

$${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$$

$$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$$

$$\therefore$$ $$\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat k$$

$${\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat k$$

$$\therefore$$ Shortest distance

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

$$ \Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25$$

$$ \Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0$$

$$\therefore$$ Sum of values of $$\lambda = {{80} \over 5} = 16$$