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

The d.r.s. of the line $\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$ are $2,4,1$. $\therefore$ The d.r.s. of any line parallel to it are also $2,4,1$. The equation of the line passing through $\mathrm{A}(1,3,2)$ is $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{1}=\lambda$ (say) Then, any point on the line is $\mathrm{B}=(2\lambda +1,4\lambda +3,\lambda +2)$ The point $\mathrm{B}$ lies on the plane $3x+y+2z=5$. $$\begin{array}{ll}\therefore & 3(2\lambda +1)+4\lambda +3+2(\lambda +2)=5\\ & \Rightarrow 12\lambda +10=5\\ & \Rightarrow \lambda =\frac{-5}{12}\\ \therefore & B=\left(\frac{1}{6},\frac{4}{3},\frac{19}{12}\right)\end{array}$$