If $\overline{a},\overline{b},\overline{c}$  are non coplanar find the point of intersection of the straight line passing through the points

$2\overline{a}+3\overline{b}-c,3\overline{a}+4\overline{b}-2c$ with the line through the points  $\overline{a}-2\overline{b}+3\overline{c},\overline{a}-6\overline{b}+6\overline{c}$.

The vector eq of the line passing through the points  $2\overline{a}+3\overline{b}-\overline{c},3\overline{a}+4\overline{b}-2c$ is $\overline{r}=(1-t)(2\overline{a}+3\overline{b}-\overline{c})+t(3\overline{a}+4\overline{b}-2\overline{c})$ 

$\overline{r}=\overline{a}(2-2t+3t)+\overline{b}(3-3t+4t)+\overline{c}(-1+t-2t)$

$=\overline{a}(2+t)+\overline{b}(3+t)+\overline{c}(-1-t).....\dots \left(1\right)$

The vector equation of the plane passing through the points  $\overline{a}-2\overline{b}+3\overline{c},\overline{a}-6\overline{b}+6\overline{c}$ is 

$\overline{r}=(1-s)(\overline{a}-2\overline{b}+3\overline{c})+s(\overline{a}-6\overline{b}+6\overline{c})$

$r=\overline{a}(1-s+s)+\overline{b}(-2+2s-6s)+\overline{c}(3-3s+6s)$

$=\overline{a}+\overline{b}(-4s-2)+\overline{c}(3s+3)\dots ....\left(2\right)$

$\left(1\right)=\left(2\right)$

$2+t =1;$ $-4s-2=3+t$

$t =-1$  $3s+3=-1-\mathrm{t}-4s-2=3-1-4s=4$

$s=-1$ from $\left(2\right)$ point of intersection is 

$\overline{r}=\overline{a}+\overline{b}(4-2)+\overline{c}(-3+3)$

$\overline{r}=\overline{a}+2\overline{b}$