Question

Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2x − y + 3z − 1 = 0 and passing through (1, −2, 3).

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ x-3y+2z-5+\lambda \left(2x-y+3z-1\right)=0...\left(1\right) \\ \text{This passes through (1, -2, 3). So,} \\ 1+6+6-5+\lambda \left(2+2+9-1\right) \\ \Rightarrow 8+12\lambda =0 \\ \Rightarrow \lambda =\frac{-2}{3} \\ \text{Substituting this in (1), we get} \\ x-3y+2z-5-\frac{2}{3}\left(2x-y+3z-1\right)=0 \\ \Rightarrow -x-7y-13=0 \\ \Rightarrow x+7y+13=0 \\ \Rightarrow \overrightarrow{r}.\left(\hat{i}+7\hat{j}\right)+13=0,\text{which is the required vector equation of the plane.} \end{gathered}$$