Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let the equation of plane be, $$a(x-1)+b(y+2)+c(z-1)=0$$ which is perpendicular to $2x-2y+z=0$ and $x-y+2z=4$ $$\begin{array}{llrl}\Rightarrow & 2a-2b+c& =0\text{ and }a-b+2c=0& \\ \text{ or }& \frac{a}{-2}1& =\frac{b}{2}=\frac{c}{-2}& \\ \Rightarrow & -12& 1& -1\\ \Rightarrow & \frac{a}{-3}& =\frac{b}{-3}=\frac{c}{0}& \\ & \frac{a}{1}=\frac{b}{1}=\frac{c}{0}& & \end{array}$$ So, the equation of plane is, $x-1+y+2=0$ or $x+y+1=0$, its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2\sqrt{2}$.