Line and Plane — Maths STD 12 Science — Question

The vector equation of the plane passing through the points
$A(\bar{a}), B(\bar{b})$ and $C(\bar{c})$
$\bar{r} \cdot(\overline{A B} \times \overline{A C})=\bar{a}(\overline{A B} \times \overline{A C})$......(1)
Let $\bar{a}=\hat{i}+\hat{j}-2 \widehat{k}, \bar{b}=\hat{i}+2 \hat{j}+\widehat{k}, \bar{c}=2 \hat{i}-\hat{j}+\widehat{k}$
$\therefore \overline{A B}=\bar{b}-\bar{a}=(\hat{i}+2 \hat{j}+\widehat{k})-(\hat{i}+\hat{j}-2 \widehat{k})=\hat{j}+3 \widehat{k}$
and $\overline{A C}=\bar{c}-\bar{a}=(2 \hat{i}-\hat{j}+\widehat{k})-(\hat{i}-\hat{j}-2 \widehat{k})=\hat{i}-2 \hat{j}+3 \widehat{k}$
$\therefore \overline{A B} \times \overline{A C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 3 \\ 1 & -2 & 3\end{array}\right|$
$=(3+6) \hat{i}-(0-3) \hat{j}+(0-1) \widehat{k}$
$=9 \hat{i}+3 \hat{j}-\widehat{k}$
$\bar{a} \cdot(\overline{A B} \times \overline{A C})=(\hat{i}+\hat{j}-2 \widehat{k}) \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})$
$=1(9)+1(3)+(-2)(-1)$
$=9+3+2=14$
from (1), the vector equation of the required plane is
$\bar{r} \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
$(x \hat{i}+y \hat{j}+z \widehat{k})(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
∴ the cartesian equation of the plane is
$9 x+3 y-z=14$