Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of plane passing through $(1,-1,2)$ is $\mathrm{a}(x-1)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}-2)=0$ Since plane (i) is perpendicular to the planes $x+2y-2z=4$ and $3x+2y+z=6$ $$\begin{array}{ll}\therefore & a+2b-2c=0\\ & \text{ and }3a+2b+c=0\\ & \Rightarrow \frac{a}{6}=\frac{b}{-7}=\frac{c}{-4}\end{array}$$ $\therefore$ The equation of the required plane is $$\begin{array}{rl}& 6(x-1)-7(y+1)-4(z-2)=0\\ & \Rightarrow 6x-7y-4z-5=0\end{array}$$