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

Let the direction ratios of the line be $⟨a,b,c⟩$. It lies in $\pi$, so $2a-b+c=0$, and it is perpendicular to $l$, so $a-b-2c=0$. Hence $\frac{a}{3}=\frac{b}{5}=\frac{c}{-1}$. A point of $l$ is $(2+\lambda ,2-\lambda ,3-2\lambda )$. It lies on $\pi$ if $2(2+\lambda )-(2-\lambda )+(3-2\lambda )=4$, so $\lambda =-1$ and the point of intersection is (1,3,5). The required line is $\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-5}{-1}$.