Find the shortest distance between the lines gives by
$\overrightarrow {\rm{r}} = (8 + 3\lambda )\widehat {\rm{i}} - (9 + 16\lambda )\widehat {\rm{j}} + (10 + 7\lambda )\widehat {\rm{k}}$

and $\overrightarrow {\rm{r}} = 15\widehat {\rm{i}} + 29\widehat {\rm{j}} + 5\widehat {\rm{k}} + \mu (3\widehat {\rm{i}} + 8\widehat {\rm{j}} - 5\widehat {\rm{k}})$

It is given that, $\vec r = (8 + 3\lambda )\widehat {\rm{i}} - (9 + 16\lambda )\widehat {\rm{j}} + (10 + 7\lambda )\widehat {\rm{k}})$

$= 8\widehat {\rm{i}} - 9\widehat {\rm{j}} + 10\widehat {\rm{k}} + 3\lambda \widehat {\rm{i}} - 16\lambda \widehat {\rm{j}} + 7\lambda \widehat {\rm{k}}$

$= 8\widehat {\rm{i}} - 9\widehat {\rm{j}} + 10\widehat {\rm{k}} + \lambda (3\widehat {\rm{i}} - 16\widehat {\rm{j}} + 7\widehat {\rm{k}})$

$\Rightarrow$ $\overrightarrow {{{\rm{a}}_1}} = 8\widehat {\rm{i}} - 9\widehat {\rm{j}} + 10\widehat {\rm{k}}$

and $\overrightarrow {{{\rm{b}}_1}} = 3\widehat {\rm{i}} - 16\widehat {\rm{j}} + 7\widehat {\rm{k}}$

…..(i)
Also $\overrightarrow {\rm{r}} = 15\widehat {\rm{i}} + 29\widehat {\rm{j}} + 5\widehat {\rm{k}} + \mu (3\widehat {\rm{i}} + 8\widehat {\rm{j}} - 5\widehat {\rm{k}})$

$\Rightarrow$ $\overrightarrow {{{\rm{a}}_2}} = 15\widehat {\rm{i}} + 29\widehat {\rm{j}} + 5\widehat {\rm{k}}$

and $\overrightarrow {{{\rm{b}}_2}} = 3\widehat {\rm{i}} + 8\widehat {\rm{j}} - 5\widehat {\rm{k}}$

……(ii)
Now, shortest distance between two lines is given by $\left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|$

$\therefore \overrightarrow {{{\rm{b}}_1}} \times \overrightarrow {{{\rm{b}}_2}}$

$= \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\3&{ - 16}&7\\3&8&{ - 5}\end{array}} \right|$

$= \widehat {\rm{i}}(80 - 56) - \widehat {\rm{j}}( - 15 - 21) + \widehat {\rm{k}}(24 + 48)$

$= 24\widehat {\rm{i}} + 36\widehat {\rm{j}} + 72\widehat {\rm{k}}$

Now, $\left| {\overrightarrow {{{\rm{b}}_1}} \times \overrightarrow {{{\rm{b}}_2}} } \right| = \sqrt {{{(24)}^2} + {{(36)}^2} + {{(72)}^2}}$

$= 12\sqrt {{2^2} + {3^2} + {6^2}} = 84$

and $\left( {\overrightarrow {{{\rm{a}}_2}} - \overrightarrow {{{\rm{a}}_1}} } \right) = (15 - 8)\widehat {\rm{i}} + (29 + 9)\widehat {\rm{j}} + (5 - 10)\widehat {\rm{k}}$.

$= 7\widehat {\rm{i}} + 38\widehat {\rm{j}} - 5\widehat {\rm{k}}$

$\therefore$ Shortest distance $= \mid \left| {\frac{{(24\widehat {\rm{i}} + 36\widehat {\rm{j}} + 72\widehat {\rm{k}}) \cdot (7\widehat {\rm{i}} + 38\widehat {\rm{j}} - 5\widehat {\rm{k}})}}{{84}}} \right|$

$= \left| {\frac{{168 + 1368 - 360}}{{84}}} \right| = \left| {\frac{{1176}}{{84}}} \right| = 14$ units