Show that the line joining the points  $6\overline{a}-4\overline{b}+4\overline{c},-4\overline{c}$  and the line joining the pair  of points,  $-\overline{a}-2\overline{b}+3\overline{c},\overline{a}+2\overline{b}-5\overline{c}$ intersect at the point $-4\overline{c}$ when $\overline{a},\overline{b},\overline{c}$  are non coplanar vectors.

let  $\overline{OA}=6\overline{a}-4\overline{b}+4\overline{c},\overline{OB}=-4c$, $\overline{OC}=-\overline{a}-2\overline{b}-3\overline{c}$,$\overline{OD}=\overline{a}+2\overline{b}-5\overline{c}$

The vector equation of the line joining whose position vectors are  $\overline{OA}$ and $\overline{OB}$ is $\overline{r}=(1-t)\overline{OA}+t\overline{OB}$ where $t\in R$

$\overline{r}=(1-t)(6\overline{a}-4\overline{b}+4\overline{c})+t(-4\overline{c}) \dots \dots .\left(1\right)$

The vector equation of the line joining the points whose position vectors are $\overline{OC}$ and $\overline{OD}$ is

$\overline{r}=(1-s)\overline{OC}+s\overline{ OD}$  where  $s\in R$$\overline{r}=(1-s)(-\overline{a}-2\overline{b}-3\overline{c})+s(\overline{a}+2\overline{b}-5\overline{c})\dots ....\left(2\right)$

From (1) & (2)

$(1-t)(6\overline{a}-4\overline{b}+4\overline{c})+t(-4\overline{c})=$

$(1-s)(-\overline{a}-2\overline{b}-3\overline{c})+s(\overline{a}+2\overline{b}-5\overline{c})$

Equating the coefficients of $\overline{a}$ and $\overline{b}$

$\Rightarrow 6-6t=-1+s+s$

$\Rightarrow 6t+2s-7=0\dots ..\left(3\right)$

and $-4+4t=-2+2s+2s$

$\Rightarrow 4t-4s-2=0$

$\Rightarrow 2t-2s-1\mid =0\dots \dots \left(4\right)$

$\left(3\right)+\left(4\right)\Rightarrow 8t=8,\Rightarrow t=1$

Point of intersection of $\left(1\right) \& \left(2\right)$ is

$(1-1)(6\overline{a}-4\overline{b}+4\overline{c})+1(-4\overline{c})=-4\overline{c}$