Three Dimensional Geometry — Class 12 Maths Solution

As we know, vector equation of a line passes through two points is represented by $\overrightarrow {\rm{r}} = \overrightarrow {\rm{a}} + \lambda (\overrightarrow {\rm{b}} - \overrightarrow {\rm{a}} )$

Here, $\overrightarrow {\rm{r}} = x\widehat {\rm{i}} + y\widehat {\rm{j}} + 3\widehat {\rm{k}},$ $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} + 4\widehat {\rm{j}} - 7\widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = \widehat {\rm{i}} - \widehat {\rm{j}} + 6\widehat {\rm{k}}$

$\Rightarrow$ $(\overrightarrow {\rm{b}} - \overrightarrow {\rm{a}} ) = - 2\widehat {\rm{i}} - 5\widehat {\rm{j}} + 13\widehat {\rm{k}}$

So, the required equation is $x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}} = 3\widehat {\rm{i}} + 4\widehat {\rm{j}} - 7\widehat {\rm{k}} + \lambda ( - 2\widehat {\rm{i}} - 5\widehat {\rm{j}} + 13\widehat {\rm{k}})$

$\Rightarrow$ $(x - 3)\widehat {\rm{i}} + (y - 4)\widehat {\rm{j}} + (z + 7)\widehat {\rm{k}} = \lambda ( - 2\widehat {\rm{i}} - 5\widehat {\rm{j}} + 13\widehat {\rm{k}})$