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

Equation of the plane passing through the intersection line of given planes is $\begin{array}{ll}& (x+2y+3z-2)+\lambda (x-y+z-3)=0\\ \text{ or }& (1+\lambda )x+(2-\lambda )y+(3+\lambda )z+(-2-3\lambda )=0\end{array}$ $\because$ Its distance from the point $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$ $\begin{array}{l}\therefore \left|\frac{3(1+\lambda )+1(2-\lambda )-1(3+\lambda )+(-2-3\lambda )}{\sqrt{(1+\lambda {)}^2+(2-\lambda {)}^2+(3+\lambda {)}^2}}\right|=\frac{2}{\sqrt{3}}\\ \Rightarrow \left|\frac{-2\lambda }{\sqrt{3{\lambda }^2+4\lambda +14}}\right|=\frac{2}{\sqrt{3}}\\ \Rightarrow 3{\lambda }^2+4\lambda +14=3{\lambda }^2\Rightarrow \lambda =-\frac{7}{2}\end{array}$ $\therefore$ Required equation of plane is $(x+2y+3z-2)-\frac{7}{2}(x-y+z-3)=0$ or $5x-11y+z=17$