Question

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line

Answer 1

$$\begin{gathered} \text{The equation of the plane through (0,0, 0) is} \\ a\left(x-0\right)+b\left(y-0\right)+c\left(z-0\right)=0 \\ ax+by+cz=0...\left(1\right) \\ \text{This plane passes through (3, -1, 2). So,} \\ 3a-b+2c=0...\left(2\right) \\ \text{Again plane (1) is parallel to the given line.} \\ \text{It means that the normal to plane (1) is perpendicular to the line.} \\ \Rightarrow a\left(1\right)+b\left(-4\right)+c\left(7\right)=0...\left(3\right)\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x& y& z\\ 3& -1& 2\\ 1& -4& 7\end{array}\right|=0 \\ \Rightarrow x-19y-11z=0 \end{gathered}$$