Find the vector equation of the plane passing through the points  $4\overline{i}-3\overline{j}-\overline{k},3\overline{i}+7\overline{j}-10\overline{k}$ and $2\overline{i}+5\overline{i}-7\overline{k}$ and show that the point  $\overline{i}+2\overline{j}-3\overline{k}$  lies in the plane

Let $\overline{OA}=4\overline{i}-3\overline{j}-\overline{k},\overline{ OB}=3\overline{i}+7\overline{i}-10\overline{k}$

$\overline{OC}=2\overline{i}+5\overline{j}-7\overline{k},\overline{OD}=\overline{i}+2\overline{i}-3\overline{k}$

The vector equation of the plane passing through the points whose position vectors $\overline{OA},\overline{OB},\overline{OC}$

is $\overline{r}=(1-s-t)\overline{OA}+s\overline{OB}+t\overline{OC}$  Where  $s,t\in R$

$\Rightarrow \overline{r}=(1-s-t)(4\overline{i}-3\overline{j}-\overline{k})$$+s(3\overline{i}+7\overline{j}-10\overline{k})+t(2\overline{i}+5\overline{j}-7\overline{k})$

$\overline{AB}=\overline{OB}-\overline{OA}=-\overline{i}+10\overline{j}-9\overline{k}$

$\overline{AC}=\overline{OC}-OA=-2\overline{i}+8\overline{j}-6\overline{k}$

$\overline{AD}=\overline{OD}-\overline{OA}=-3\overline{i}+5\overline{j}-2\overline{k}$

$[\overline{AB} \overline{AC} \overline{AD}]=\left|\begin{array}{ccc}-1& 10& -9\\ -2& 8& -6\\ -3& 5& -2\end{array}\right|$

$=[-(14)-10(-14)-9(14\left)\right]$ 

$=(-14+140-126)$

$=[140-140]=0$

$\therefore \overline{AB},\overline{AC},\overline{AD}$ collinear are vectors.

$\therefore$The given point lie on the same plane