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

Given $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \dots$ (say) ...(1) Hence coordinates of any point on this line are $\therefore x=\lambda +3,y=2\lambda +4,z=2\lambda +5$ Since this point lies on the plane, we write $\begin{array}{l}(\lambda +3)+(2\lambda +4)+(2\lambda +5)=2\\ 5\lambda +12=2\Rightarrow \lambda =-2\end{array}$ Hence coordinates of point of inter section are $\equiv (-2+3,-4+4,-4+5)$ i.e. $(1,0,1)$ $\therefore$ Distance between $(3,4,5)$ and $(1,0,1)$ is $=\sqrt{(3-1{)}^2+(4-0{)}^2+(5-1{)}^2}=6$