Question

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

Answer 1

$$\begin{gathered} \text{The equation of the plane through (2, 2, -1) is} \\ a\left(x-2\right)+b\left(y-2\right)+c\left(z+1\right)=0...\left(1\right) \\ \text{This plane passes through (3, 4, 2). So,} \\ a\left(3-2\right)+b\left(4-2\right)+c\left(2+1\right)=0 \\ \Rightarrow a+2b+3c=0...\left(2\right) \\ \text{Again plane (1) is parallel to the line whose direction ratios are 7, 0, 6.} \\ \text{It means that the normal of plane (1) is perpendicular to the}\mathrm{line\; whose\; direction\; ratios\; are\; 7,\; 0,\; 6}. \\ \Rightarrow 7a+0b+6c=0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-2& y-2& z+1\\ 1& 2& 3\\ 7& 0& 6\end{array}\right|=0 \\ \Rightarrow 12\left(x-2\right)+15\left(y-2\right)-14\left(z+1\right)=0 \\ \Rightarrow 12x+15y-14z-68=0 \end{gathered}$$