Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

(B) The equation of the plane parallel to the plane $x+2y+2z+18=0$ is $x+2y+2z+\lambda =0$. Now, the distance of this plane from the print $(1,1,2)$ is $\therefore \left|\frac{1(1)+2(1)+2(2)+\lambda }{\sqrt{1^2+2^2+2^2}}\right|=\left|\frac{7+\lambda }{3}\right|$ $\begin{array}{l}\text{ We have }\left|\frac{7+\lambda }{3}\right|=2\Rightarrow \frac{7+\lambda }{3}=\pm 2\\ \therefore \lambda =\pm 6-7=1\Rightarrow \lambda =-1\text{ or }\lambda =-13\end{array}$ Hence, equations of plane are $x+2y+2z-1=0$ or $x+2y+2z-13=0$